# Solved Examples on Logarithmic Equations

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASSCE is a question for the WASSCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve each logarithmic equation.
Unless specified otherwise, give the exact solutions.
Show all work.
Use at least two methods as necessary.
Check your solutions as applicable.
Depending on time, it is highly recommended you check your work even if the question does not require it.
By checking your work, you can tell whether your answer is correct or incorrect [if the Left Hand Side (LHS) is equal or not equal to the Right Hand Side (RHS)].
Please make sure you check with the original equation rather than the modified equation. This is because the modified equation could be wrong.

(1.) $\log_x{128} = 7$

First Method: By Exponents

$\log_x{128} = 7 \\[3ex] x^7 = 128 ...Relationship \\[3ex] x^7 = 2^7 \\[3ex] Exponents\:\; are\:\; the\:\; same \\[3ex] Equate\: the\:\; bases \\[3ex] x = 2 \\[3ex]$ Second Method: By Logarithms

$\log_x{128} = 7 \\[3ex] 7 = \log_x{x^7} ...Laws\: 4\: and\: 5...Log \\[3ex] \rightarrow \log_x{128} = \log_x{x^7} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] 128 = x^7 \\[3ex] x^7 = 128 \\[3ex] x^7 = 2^7 \\[3ex] Exponents\: are\:\; the\:\; same \\[3ex] Equate\:\; the\:\; bases \\[3ex] x = 2 \\[3ex]$ Check
 LHS $\log_x{128} \\[3ex] \log_2{128} \\[3ex] = \log_2{2^7} = 7\log_2{2} ...Law\: 5...Log \\[3ex] \log_2{2} = 1...Law\: 4...Log \\[3ex] = 7 * 1 = 7$ RHS $7$
(2.) $\log_x{125} = -3$

First Method: By Exponents

$\log_x{125} = -3 \\[3ex] x^{-3} = 125 ...Relationship \\[3ex] x^{-3} = \dfrac{1}{x^3}...Law\: 6...Exp \\[3ex] \rightarrow \dfrac{1}{x^3} = 125 \\[3ex] 1 = 125x^3 \\[3ex] 125x^3 = 1 \\[3ex] x^3 = \dfrac{1}{125} \\[3ex] x = \sqrt[3]{\dfrac{1}{125}} \\[3ex] x = \dfrac{1}{5} \\[5ex]$ Second Method: By Logarithms

$\log_x{125} = -3 \\[3ex] -3 = \log_x{x^{-3}} ...Laws\: 4\: and\: 5...Log \\[3ex] \rightarrow \log_x{125} = \log_x{x^{-3}} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] 125 = x^{-3} \\[3ex] x^{-3} = 125 \\[3ex] x^{-3} = \dfrac{1}{x^3}...Law\: 6...Exp \\[3ex] \rightarrow \dfrac{1}{x^3} = 125 \\[3ex] 1 = 125x^3 \\[3ex] 125x^3 = 1 \\[3ex] x^3 = \dfrac{1}{125} \\[3ex] x = \sqrt[3]{\dfrac{1}{125}} \\[3ex] x = \dfrac{1}{5} \\[5ex]$ Check
 LHS $\log_x{125} \\[3ex] \log_{\dfrac{1}{5}}{125} \\[3ex] = \dfrac{\log_5{125}}{\log_5{\dfrac{1}{5}}} = ...Law\: 6...Log \\[3ex] = \log_5{125} \div \log_5{\dfrac{1}{5}} \\[3ex] \log_5{125} = \log_5{5^3} = 3\log_5{5} ...Law\: 5...Log \\[3ex] \log_5{5} = 1...Law\: 4...Log \\[3ex] \log_5{125} = 3 * 1 = 3 \\[3ex] \dfrac{1}{5} = 5^{-1} ...Law\: 6...Exp \\[3ex] \log_5{\dfrac{1}{5}} = \log_5{5^{-1}} = -1\log_5{5} ...Law\: 5...Log \\[3ex] \log_5{\dfrac{1}{5}} = -1 * 1 = -1 \\[3ex] = 3 \div -1 \\[3ex] = -3$ RHS $-3$
(3.) $\log_x{2} = \dfrac{1}{4}$

First Method: By Exponents

$\log_x{2} = \dfrac{1}{4} \\[3ex] x^{\dfrac{1}{4}} = 2 ...Relationship \\[3ex] Multiply\: both\: exponents\: by\: 4 \\[3ex] x^{\dfrac{1}{4} * 4} = 2^4 \\[3ex] x = 16$
Second Method: By Logarithms

$\log_x{2} = \dfrac{1}{4} \\[3ex] \dfrac{1}{4} = \log_x{x^{\dfrac{1}{4}}} ...Laws\: 4\: and\: 5...Log \\[3ex] \rightarrow \log_x{2} = \log_x{x^{\dfrac{1}{4}}} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] 2 = x^{\dfrac{1}{4}} \\[3ex] x^{\dfrac{1}{4}} = 2 \\[3ex] Multiply\: both\: exponents\: by\: 4 \\[3ex] x^{\dfrac{1}{4} * 4} = 2^4 \\[3ex] x = 16$
Check
 LHS $\log_x{2} \\[3ex] \log_16{2} \\[3ex] = \dfrac{\log_2{2}}{\log_2{16}} ...Law\: 6...Log \\[3ex] \log_2{2} = 1...Law\: 4...Log \\[3ex] \log_2{16} = \log_2{2^4} = 4\log_2{2} ...Law\: 5...Log \\[3ex] \log_2{16} = 4 * 1 = 4 \\[3ex] = \dfrac{1}{4}$ RHS $\dfrac{1}{4}$
(4.) $\log_x{\dfrac{1}{25}} = -2$

First Method: By Exponents

$\log_x{\dfrac{1}{25}} = -2 \\[3ex] x^{-2} = \dfrac{1}{25} ...Relationship \\[3ex] x^{-2} = 25^{-1}...Law\: 6...Exp \\[3ex] Multiply\: both\: exponents\: by\: -\dfrac{1}{2} \\[5ex] x^{\left(-2 * -\dfrac{1}{2}\right)} = 25^{\left(-1 * -\dfrac{1}{2}\right)}...Law\: 5...Exp \\[5ex] x = 25^{\dfrac{1}{2}} \\[5ex] 25^{\dfrac{1}{2}} = \sqrt{25} ...Law\: 7...Exp \\[5ex] \rightarrow x = \sqrt{25} \\[3ex] x = 5$
Second Method: By Logarithms

$\log_x{\dfrac{1}{25}} = -2 \\[3ex] -2 = \log_x{x^{-2}} ...Laws\: 4\: and\: 5...Log \\[5ex] \rightarrow \log_x{\dfrac{1}{25}} = \log_x{x^{-2}} \\[5ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[5ex] \dfrac{1}{25} = x^{-2} \\[5ex] x^{-2} = \dfrac{1}{25} \\[5ex] x^{-2} = \dfrac{1}{5^2} \\[5ex] \dfrac{1}{5^2} = 5^{-2}...Law\: 6...Exp \\[5ex] \rightarrow x^{-2} = 5^{-2} \\[5ex] Exponents\: are\: the\: same \\[3ex] Equate\: the\: base \\[3ex] x = 5$
Check
 LHS $\log_x{\dfrac{1}{25}} \\[5ex] = \log_5{\dfrac{1}{25}} \\[5ex] \dfrac{1}{25} = 25^{-1} ...Law\: 6...Exp \\[5ex] 25^{-1} = 5^2{-1} \\[5ex] 5^2{-1} = 5^{(2 * -1)} = 5^{-2} ...Law\: 5...Exp \\[5ex] = \log_5{5^{-2}} \\[5ex] = -2\log_5{5} ...Law\: 5...Log \\[5ex] \log_5{5} = 1...Law\: 4...Log \\[3ex] = -2 * 1 \\[3ex] = -2$ RHS $-2$
(5.) $\log x + \log(x + 3) = 1$

$\log x + \log(x + 3) = 1 \\[3ex] \log x + \log(x + 3) = \log{[x(x + 3)]} ...Law\: 1...Log \\[3ex] 1 = \log 10 ...Law\: 4...Log \\[3ex] \rightarrow \log{[x(x + 3)]} = \log 10 \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] x(x + 3) = 10 \\[3ex] x^2 + 3x - 10 = 0 \\[3ex] (x + 5)(x - 2) = 0 \\[3ex] x + 5 = 0\:\:\: OR\:\:\: x - 2 = 0 ...Zero\: Product\: Property \\[3ex] x = -5\:\:\: OR\:\:\: x = 2 \\[3ex] Log\: of\: a\: negative\: number\: DNE ...based\: on\: the\: LHS \\[3ex] \therefore x = 2$
Check
 LHS $\log x + \log(x + 3) \\[3ex] x = 2 \\[3ex] = \log 2 + \log(2 + 3) \\[3ex] = \log 2 + \log 5 \\[3ex] = \log (2 * 5) ...Law\: 1...Log \\[3ex] = \log 10 \\[3ex] = 1 ...Law\: 4...Log$ RHS $1$
(6.) $(\log_3{x})^2 - 2\log_3{x} = 15$

$(\log_3{x})^2 - 2\log_3{x} = 15 \\[5ex] Let \log_3{x} = p \\[3ex] \rightarrow p^2 - 2p = 15 \\[3ex] p^2 - 2p - 15 = 0 \\[3ex] (p + 3)(p - 5) = 0 \\[3ex] p + 3 = 0\:\:\: OR\:\:\: p - 5 = 0 ...Zero\: Product\: Property \\[3ex] p = -3\:\:\: OR\:\:\: p = 5 \\[3ex] Recall: \log_3{x} = p \\[3ex] When\: p = -3 \\[3ex] \log_3{x} = -3 \\[3ex] x = 3^{-3} ...Relationship \\[3ex] 3^{-3} = \dfrac{1}{3^3} ...Law\: 6...Exp \\[5ex] \rightarrow x = \dfrac{1}{3^3} \\[5ex] x = \dfrac{1}{27} \\[7ex] When\: p = 5 \\[3ex] \log_3{x} = 5 \\[3ex] x = 3^{5} ...Relationship \\[3ex] x = 243$
Check
 LHS $(\log_3{x})^2 - 2\log_3{x} \\[5ex] x = \dfrac{1}{27} \\[5ex] = (\log_3{\dfrac{1}{27}})^2 - 2\log_3{\dfrac{1}{27}} \\[5ex] \log_3{\dfrac{1}{27}} = \log_3{27^{-1}} ...Law\: 6...Exp \\[5ex] \log_3{27^{-1}} = \log_3{3^{3^({-1})}} \\[5ex] \log_3{3^{3^({-1})}} = \log_3{3^{-3}} ...Law\: 5...Exp \\[5ex] \log_3{3^{-3}} = -3\log_3{3} ...Law\: 5...Log \\[5ex] -3\log_3{3} = -3 * 1 ...Law\: 3...Log \\[5ex] -3 * 1 = -3 \\[5ex] = (-3)^2 - 2(-3) \\[5ex] = 9 + 6 \\[3ex] = 15$ $(\log_3{x})^2 - 2\log_3{x} \\[3ex] x = 243 \\[3ex] (\log_3{243})^2 - 2\log_3{243} \\[3ex] \log_3{243} = \log_3{3^5} \\[3ex] \log_3{3^5} = 5\log_3{3} ...Law\: 5...Log \\[3ex] 5\log_3{3} = 5 * 1 ...Law\: 3...Log \\[5ex] 5 * 1 = 5 \\[5ex] = (5)^2 - 2(5) \\[5ex] = 25 - 10 \\[3ex] = 15$ RHS $15$
(7.) $\log_4{(x - 5)} + \log_4{(x + 1)} = 2$

First Method
$\log_4{(x - 5)} + \log_4{(x + 1)} = 2 \\[3ex] \log_4{(x - 5)} + \log_4{(x + 1)} = \log_4{[(x - 5)(x + 1)]} ...Law\: 1...Log \\[3ex] 2 = \log_4{4^2} ...Laws\: 4\: and\: 5...Log \\[3ex] 2 = \log_4{16} \\[3ex] \rightarrow \log_4{[(x - 5)(x + 1)]} = \log_4{16} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] (x - 5)(x + 1) = 16 \\[3ex] x^2 + x - 5x - 5 = 16 \\[3ex] x^2 - 4x - 5 - 16 = 0 \\[3ex] x^2 - 4x - 21 = 0 \\[3ex] (x + 3)(x - 7) = 0 \\[3ex] x + 3 = 0\:\:\: OR\:\:\: x - 7 = 0 ...Zero\: Product\: Property \\[3ex] x = -3\:\:\: OR\:\:\: x = 7 \\$
Second Method
$\log_4{(x - 5)} + \log_4{(x + 1)} = 2 \\[3ex] \log_4{(x - 5)} + \log_4{(x + 1)} = \log_4{[(x - 5)(x + 1)]} ...Law\: 1...Log \\[3ex] \rightarrow \log_4{[(x - 5)(x + 1)]} = 2 \\[3ex] 4^2 = [(x - 5)(x + 1)] ...Relationship \\[3ex] 16 = (x - 5)(x + 1) \\[3ex] (x - 5)(x + 1) = 16 \\[3ex] x^2 + x - 5x - 5 = 16 \\[3ex] x^2 - 4x - 5 - 16 = 0 \\[3ex] x^2 - 4x - 21 = 0 \\[3ex] (x + 3)(x - 7) = 0 \\[3ex] x + 3 = 0\:\:\: OR\:\:\: x - 7 = 0 ...Zero\: Product\: Property \\[3ex] x = -3\:\:\: OR\:\:\: x = 7 \\$
Check
 LHS $\log_4{(x - 5)} + \log_4{(x + 1)} \\[3ex] x = -3 \\[3ex] = \log_4{(-3 - 5)} + \log_4{(-3 + 1)} \\[3ex] = \log_4{-8} + \log_4{-2} \\[3ex] = DNE \\$ $x = -3$ is not a solution. $\log_4{(x - 5)} + \log_4{(x + 1)} \\[3ex] x = 7 \\[3ex] = \log_4{(7 - 5)} + \log_4{(7 + 1)} \\[3ex] = \log_4{2} + \log_4{8} \\[3ex] = \log_4{(2 * 8)} \\[3ex] = \log_4{16} \\[3ex] = log_4{4^2} = 2\log_4{4} ...Law\: 5...Log \\[3ex] \log_4{4} = 1...Law\: 4...Log \\[3ex] = 2 * 1 \\[3ex] = 2$ RHS $2$
(8.) $7$$\log_7{10} = 5x 7$$\log_7{10}$ = $5x$

$7$$\log_7{10} = 10 ...Law\: 7...Log \implies \\[3ex] 10 = 5x \\[3ex] 5x = 10 \\[3ex] x = 2 Check  LHS 7$$\log_7{10}$ $7$$\log_7{10}$ = $10 ...Law\: 7...Log$ $10$ RHS $5x \\[3ex] x = 2 \\[3ex] 5(2) \\[3ex] 10$
(9.) ACT When $\log_5{x} = -2$, what is $x$?

$F.\:\: -32 \\[3ex] G.\:\: -25 \\[3ex] H.\:\: -10 \\[3ex] J.\:\: \dfrac{1}{10} \\[5ex] K.\:\: \dfrac{1}{25} \\[5ex]$

First Method: By Exponents

$\log_5{x} = -2 \\[3ex] 5^{-2} = x ...Relationship \\[3ex] x = 5^{-2} = \dfrac{1}{5^2} ...Law\:\: 6...Exp \\[5ex] x = \dfrac{1}{25} \\[5ex]$ Second Method: By Logarithms

$\log_5{x} = -2 \\[3ex] -2 = \log_5{5^{-2}} ...Laws\: 4\: and\: 5...Log \\[3ex] \rightarrow \log_5{x} = \log_5{5^{-2}} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] x = 5^{-2} = \dfrac{1}{5^2} ...Law\:\: 6...Exp \\[5ex] x = \dfrac{1}{25} \\[5ex]$ Check
 LHS $\log_5{x} \\[3ex] x = \dfrac{1}{25} \\[5ex] \log_5{\dfrac{1}{25}} \\[3ex] \dfrac{1}{25} = \dfrac{1}{5^2} = 5^{-2} ...Law\:\: 6...Exp \\[5ex] = log_5{5^{-2}} ...Law\: 6...Log \\[3ex] = -2 * \log_5{5} ...Law\:\: 5...Log \\[3ex] \log_5{5} = 1...Law\: 4...Log \\[3ex] = -2 * 1 \\[3ex] = -2$ RHS $-2$
(10.) ACT Which of the following values is the $x-coordinate$ of the point in the standard $(x, y)$ coordinate plane where the graph of the line $y = 7$ intersects the graph of the function $y = \ln(x - 2) + 3$?

$A.\:\: 6 \\[3ex] B.\:\: e^4 + 2 \\[3ex] C.\:\: 4e + 2 \\[3ex] D.\:\: \ln(4) + 2 \\[3ex] E.\:\: \ln(5) + 3 \\[3ex]$

"Intersect" means that the two lines "meet"
So, equate the two lines/equations

$y = y \\[3ex] 7 = \ln(x - 2) + 3 \\[3ex] \ln(x - 2) + 3 = 7 \\[3ex] \ln(x - 2) = 7 - 3 \\[3ex] \ln(x - 2) = 4 \\[3ex] \log_e{(x - 2)} = 4 \\[3ex] \implies x - 2 = e^4 \\[3ex] x = e^4 + 2$
(11.) $\ln (p + 12) + \ln (p - 5) = 2 \ln p$

$\ln (p + 12) + \ln (p - 5) = 2 \ln p \\[3ex] \ln (p + 12) + \ln (p - 5) = \ln [(p + 2)(p - 5)] ...Law\: 1...Log \\[3ex] (p + 2)(p - 5) = p^2 - 5p + 12p - 60 ...FOIL \\[3ex] p^2 - 5p + 12p - 60 = p^2 + 7p - 60 \\[3ex] \ln [(p + 2)(p - 5)] = \ln (p^2 + 7p - 60) \\[3ex] 2 \ln p = \ln p^2 ...Law\: 5...Log \\[3ex] \rightarrow \ln (p^2 + 7p - 60) = \ln p^2 \\[3ex] Same\: base \\[3ex] Equate\: the\: terms \\[3ex] p^2 + 7p - 60 = p^2 \\[3ex] p^2 - p^2 + 7p - 60 = 0 \\[3ex] 7p = 60 \\[3ex] p = \dfrac{60}{7}$
Check
 LHS $\ln (p + 12) + \ln (p - 5) \\[3ex] p = \dfrac{60}{7} \\[3ex] p + 12 = \dfrac{60}{7} + 12 \\[5ex] \dfrac{60}{7} + 12 = \dfrac{60}{7} + \dfrac{84}{7} \\[5ex] \dfrac{60}{7} + \dfrac{84}{7} = \dfrac{60 + 84}{7} \\[5ex] \dfrac{60 + 84}{7} = \dfrac{144}{7} \\[5ex] p - 5 = \dfrac{60}{7} - 5 \\[5ex] \dfrac{60}{7} - 5 = \dfrac{60}{7} - \dfrac{35}{7} \\[5ex] \dfrac{60}{7} - \dfrac{35}{7} = \dfrac{60 - 35}{7} \\[5ex] \dfrac{60 - 35}{7} = \dfrac{25}{7} \\[5ex] = \ln \dfrac{144}{7} + \ln \dfrac{25}{7} \\[5ex] = \ln \left(\dfrac{144}{7} * \dfrac{25}{7}\right) ...Law\: 1...Log \\[5ex] = \ln \dfrac{3600}{49}$ RHS $2 \ln p \\[3ex] = \ln p^2 ...Law\: 5...Log \\[3ex] p = \dfrac{60}{7} \\[5ex] = \ln \left(\dfrac{60}{7}\right)^2 \\[5ex] = \ln \left(\dfrac{60^2}{7^2}\right) \\[5ex] = \ln \dfrac{3600}{49}$
(12.) $\log_4{(p + 27)} - \log_4{(p + 7)} = \log_4{p}$

$\log_4{(p + 27)} - \log_4{(p + 7)} = \log_4{p} \\[3ex] \log_4{(p + 27)} - \log_4{(p + 7)} = \log_4{\left(\dfrac{p + 27}{p + 7}\right)} ...Law\: 2...Log \\[5ex] \rightarrow \log_4{\left(\dfrac{p + 27}{p + 7}\right)} = \log_4{p} \\[5ex] Same\: base \\[3ex] Equate\: the\: terms \\[3ex] \dfrac{p + 27}{p + 7} = p \\[3ex] LCD = p + 7 \\[3ex] Multiply\: both\: sides\: by\: the\: LCD \\[3ex] p + 27 = p(p + 7) \\[3ex] p + 27 = p^2 + 7p \\[3ex] 0 = p^2 + 7p - p - 27 \\[3ex] 0 = p^2 + 6p - 27 \\[3ex] p^2 + 6p - 27 = 0 \\[3ex] (p + 9)(p - 3) = 0 \\[3ex] p + 9 = 0\:\:\: OR\:\:\: p - 3 = 0 ...Zero\: Product\: Property \\[3ex] p = -9\:\:\: OR\:\:\: p = 3 \\[3ex] Log\: of\: a\: negative\: number\: DNE ...based\: on\: the\: RHS \\[3ex] \therefore p = 3$
Check
 LHS $\log_4{(p + 27)} - \log_4{(p + 7)} \\[3ex] = \log_4{\left(\dfrac{p + 27}{p + 7}\right)} ...Law\: 2...Log \\[5ex] p = 3 \\[3ex] p + 27 = 3 + 27 = 30 \\[3ex] p + 7 = 3 + 7 = 10 \\[3ex] = \log_4{\left(\dfrac{30}{10}\right)} \\[5ex] = \log_4{3}$ RHS $\log_4{p} \\[3ex] p = 3 \\[3ex] = \log_4{3}$
(13.) USSCE Advance Mathematics Paper 2, 2011
Solve the simultaneous equations for x and y

$\log_2 x + \log_2 y = 2 \\[3ex] \log_2 x - \log_2 y = 0 \\[3ex]$

$\log_2 x + \log_2 y = 2 \\[4ex] 2 = \log_2{2^{2}}...Law\;5...Log \\[4ex] \implies \log_2 x + \log_2 y = \log_2{2^{2}} \\[4ex] \log_2 x + \log_2 y = \log_2{xy}...Law\;1...Log \\[4ex] \implies \log_2{xy} = \log_2{2^{2}} \\[4ex] Cancel\;\;log \\[3ex] xy = 2^2 \\[3ex] xy = 4...eqn.(1) \\[3ex] Another\;\;Approach: \\[3ex] log_2 x - \log_2 y = 0 \\[4ex] \log_2 x - \log_2 y = \log_2{\dfrac{x}{y}}...Law\;2...Log \\[5ex] \implies \log_2{\dfrac{x}{y}} = 0 \\[4ex] \dfrac{x}{y} = 2^0 ... Relationship \\[5ex] 2^0 = 1...Law\;3...Exp \\[3ex] \implies \dfrac{x}{y} = 1 \\[5ex] x = 1(y) \\[3ex] x = y ...eqn.(2) \\[3ex] Subst.\;\;x\;\;for\;\;y\;\;in\;\;eqn.(1) \\[3ex] xy = 4...eqn.(1) \\[3ex] x(x) = 4 \\[3ex] x^2 = 4 \\[3ex] x = \pm \sqrt{4} \\[3ex] x = \pm 2 \\[3ex] x\;\;cannot\;\;be\;\;negative \\[3ex] \therefore x = 2 \\[3ex] From\;\;eqn.(2) \\[3ex] x = y \\[3ex] \therefore y = 2 \\[3ex]$ Check
 LHS $\log_2 x + \log_2 y \\[4ex] \log_2 2 + \log_2 2 \\[4ex] \log_2 2 = 1...Law\;4...Log \\[4ex] = 1 + 1 \\[3ex] = 2 \\[3ex]$ $\log_2 x - \log_2 y \\[4ex] \log_2 2 - \log_2 2 \\[4ex] = 1 - 1 \\[3ex] = 0$ RHS $2 \\[3ex]$ $0$
(14.) WASSCE Find the value of $y$ if

$\dfrac{\log_{10}y}{\log_{10}64} = \dfrac{1}{2}$

$\dfrac{\log_{10}y}{\log_{10}64} = \dfrac{1}{2} \\[5ex] Cross\;\;multiply \\[3ex] 2\log_{10}y = \log_{10}64 \\[3ex] 2\log_{10}y = \log_{10}y^2...Law\;5...Log \\[3ex] \rightarrow \log_{10}y^2 = \log_{10}64 \\[3ex] Same\;\;Log\;\;base \\[3ex] y^2 = 64 \\[3ex] y = \sqrt{64} \\[3ex] y = \pm 8 \\[3ex] y\;\;cannot\;\;be\;\;negative \\[3ex] \therefore y = 8 \\[3ex]$ Check
$\dfrac{\log_{10}y}{\log_{10}64} = \dfrac{1}{2}$

$y = 8$
LHS RHS
$\dfrac{\log_{10}y}{\log_{10}64} \\[5ex] = \dfrac{\log_{10}8}{\log_{10}64} \\[5ex] = \dfrac{\log_{10}8^1}{\log_{10}8^2} \\[5ex] = \dfrac{1\log_{10}8}{2\log_{10}8} \\[5ex] = \dfrac{1}{2}$ $\dfrac{1}{2}$
(15.) WASSCE Express $a \log_y{b} = x$ in index form.

$A.\;\; a^y = b^x \\[3ex] B.\;\; y^x = a^b \\[3ex] C.\;\; y^x = b^a \\[3ex] D.\;\; y^b = a^x \\[3ex]$

$a \log_y{b} = x \\[3ex] a \log_y{b} = \log_y{b^a} ...Law\;5...Log \\[3ex] \implies \\[3ex] \log_y{b^a} = x \\[3ex] y^x = b^a ...Definition$
(16.) JAMB Find x if $\log_{9}x = 1.5$

$A.\;\; 72.0 \\[3ex] B.\;\; 27.0 \\[3ex] C.\;\; 36.0 \\[3ex] D.\;\; 3.5 \\[3ex] E.\;\; 24.5 \\[3ex]$

$\log_{9}x = 1.5 \\[3ex] 9^{1.5} = x \\[3ex] x = 9^{1.5} \\[3ex] x = 9^{\dfrac{15}{10}} = 9^{\dfrac{3}{2}} \\[5ex] x = \left(\sqrt{9}\right)^3 \\[3ex] x = 3^3 \\[3ex] x = 27 \\[3ex]$ Check
$x = 27$
LHS RHS
$\log_{9}x \\[3ex] \log_{9}{27} \\[3ex] \dfrac{\log_3{27}}{\log_3{9}} \\[5ex] \dfrac{\log_3{3^3}}{\log_3{3^2}} \\[5ex] \dfrac{3\log_3{3}}{2\log_3{3}} \\[5ex] \dfrac{3(1)}{2(1)} \\[5ex] \dfrac{3}{2} \\[5ex] 1.5$ $1.5$
(17.) NYSED If $\log_3{(x + 1)} - \log_3{x} = 2$, then x equals

$(1)\;\; -\dfrac{9}{8} \hspace{10em} (3)\;\; \dfrac{1}{8} \\[5ex] (2)\;\; -\dfrac{6}{5} \hspace{10em} (4)\;\; \dfrac{1}{5} \\[5ex]$

$\log_3{(x + 1)} - \log_3{x} = 2 \\[3ex] \log_3{\left(\dfrac{x + 1}{x}\right)} = 2 \\[5ex] \dfrac{x + 1}{x} = 3^2 \\[5ex] \dfrac{x + 1}{x} = 9 \\[5ex] x + 1 = 9x \\[3ex] 9x = x + 1 \\[3ex] 9x - x = 1 \\[3ex] 8x = 1 \\[3ex] x = \dfrac{1}{8} \\[5ex]$ Check
$x = \dfrac{1}{8}$
LHS RHS
$\log_3{(x + 1)} - \log_3{x} \\[3ex] \log_3{\left(\dfrac{1}{8} + 1\right)} - \log_3{\left(\dfrac{1}{8}\right)} \\[5ex] \log_3{\left(\dfrac{1}{8} + \dfrac{8}{8}\right)} - \log_3{\left(\dfrac{1}{8}\right)} \\[5ex] \log_3{\left(\dfrac{9}{8}\right)} - \log_3{\left(\dfrac{1}{8}\right)} \\[5ex] \log_3{\left(\dfrac{9}{8} \div \dfrac{1}{8}\right)} \\[5ex] \log_3{\left(\dfrac{9}{8} * \dfrac{8}{1}\right)} \\[5ex] \log_3{9} \\[3ex] \log_3{3^2} \\[3ex] 2\log_3{3} \\[3ex] 2 * 1 \\[3ex] 2$ $2$
(18.) WASSCE If $\log_2{(3x - 1)} = 5$, find x

$A.\;\; 2.00 \\[3ex] B.\;\; 3.67 \\[3ex] C.\;\; 8.67 \\[3ex] D.\;\; 11.00 \\[3ex]$

$\log_2{(3x - 1)} = 5 \\[3ex] 2^5 = 3x - 1 \\[3ex] 32 = 3x - 1 \\[3ex] 3x - 1 = 32 \\[3ex] 3x = 32 + 1 \\[3ex] 3x = 33 \\[3ex] x = \dfrac{33}{3} \\[5ex] x = 11 \\[3ex]$ Check
$x = 11$
LHS RHS
$\log_2{(3x - 1)} \\[3ex] \log_2{[3(11) - 1]} \\[3ex] \log_2{(33 - 1)} \\[3ex] \log_2{32} \\[3ex] \log_2{2^5} \\[3ex] 5\log_2{2} \\[3ex] 5(1) \\[3ex] 5$ $5$
(19.) Use the one-to-one property of logarithms to find an exact solution for $\ln(3) + \ln(3x^2 - 5) = \ln(154)$

$\ln(3) + \ln(3x^2 - 5) = \ln(154) \\[3ex] \ln[3(3x^2 - 5)] = \ln 154...Law\;1...Log \\[3ex] Same\;\;base \\[3ex] Equate\;\;the\;\;terms \\[3ex] 3(3x^2 - 5) = 154 \\[3ex] 9x^2 - 15 = 154 \\[3ex] 9x^2 = 154 + 15 \\[3ex] 9x^2 = 169 \\[3ex] x^2 = \dfrac{169}{9} \\[5ex] x = \pm \sqrt{\dfrac{169}{9}} \\[5ex] x = \pm \dfrac{13}{3} \\[5ex]$ Check
$x = \pm \dfrac{13}{3}$
LHS RHS
$\ln(3) + \ln(3x^2 - 5) \\[3ex] When\;\;x = -\dfrac{13}{3} \\[5ex] \ln(3) + \ln\left[(3\left(-\dfrac{13}{3}\right)^2 - 5\right] \\[5ex] \ln(3) + \ln\left[(3\left(\dfrac{13}{3}\right)^2 - 5\right] \\[5ex] \ln(3) + \ln\left[3 * \dfrac{13}{3} * \dfrac{13}{3} - 5\right] \\[5ex] \ln(3) + \ln\left[\dfrac{169}{3} - 5\right] \\[5ex] \ln\left[3\left(\dfrac{169}{3} - 5\right)\right]...Law\;1...Log \\[5ex] \ln(169 - 15) \\[3ex] \ln(154)$
$When\;\;x = \dfrac{13}{3} \\[5ex] \ln(3) + \ln\left[(3\left(\dfrac{13}{3}\right)^2 - 5\right] \\[5ex] \ln(3) + \ln\left[3 * \dfrac{13}{3} * \dfrac{13}{3} - 5\right] \\[5ex] \ln(3) + \ln\left[\dfrac{169}{3} - 5\right] \\[5ex] \ln\left[3\left(\dfrac{169}{3} - 5\right)\right]...Law\;1...Log \\[5ex] \ln(169 - 15) \\[3ex] \ln(154)$
$\ln 154$
(20.) WASSCE If $(y - 1)\log_{10}4 = y\log_{10}16$, without using Mathematical tables or calculator, find the value of y

$(y - 1)\log_{10}4 = y\log_{10}16 \\[3ex] \log_{10}4^{y - 1} = \log_{10}16^y ...Law\;5...Log \\[3ex] same\;\;base;\;\;cancel\;\;log;\;\;equate\;\;terms \\[3ex] 4^{y - 1} = 16^y \\[3ex] 4^{y - 1} = 4^{2(y)} \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] y - 1 = 2y \\[3ex] 2y = y - 1 \\[3ex] 2y - y = -1 \\[3ex] y = -1 \\[3ex]$ Check
$y = -1$
LHS RHS
$(y - 1)\log_{10}4 \\[3ex] (-1 - 1)\log_{10}4 \\[3ex] -2\log_{10}4 \\[3ex] \log_{10}4^{-2} \\[3ex] \log_{10}{\left(\dfrac{1}{4^2}\right)} \\[5ex] \log_{10}{\left(\dfrac{1}{16}\right)}$ $y\log_{10}16 \\[3ex] -1\log_{10}16 \\[3ex] \log_{10}16^{-1} \\[3ex] \log_{10}{\left(\dfrac{1}{16}\right)}$

(21.) NSC Solve for x: $\log_{3}(x + 36) = \log_3{2x} + \log 100$

$\log_{3}(x + 36) = \log_3{2x} + \log 100 \\[3ex] But:\;\;\log 100 = \log_{10}100 = \log_{10}10^2 = 2\log_{10}10 = 2(1) = 2 \\[3ex] \implies \\[3ex] \log_{3}(x + 36) = \log_3{2x} + 2 \\[3ex] But:\;\;2 = \log_{3}3^2 \\[3ex] \implies \\[3ex] \log_{3}(x + 36) = \log_{3}{2x} + \log_{3}3^2 \\[3ex] \log_{3}(x + 36) = \log_{3}(2x * 3^2) \\[3ex] same\;\;base;\;\;cancel\;\;log;\;\;equate\;\;terms \\[3ex] x + 36 = 2x * 9 \\[3ex] x + 36 = 18x \\[3ex] 18x = x + 36 \\[3ex] 18x - x = 36 \\[3ex] 17x = 36 \\[3ex] x = \dfrac{36}{17} \\[5ex]$ Check
$x = \dfrac{36}{17}$
LHS RHS
$\log_{3}(x + 36) \\[3ex] \log_{3}\left(\dfrac{36}{17} + 36\right) \\[5ex] \log_{3}\left(\dfrac{36}{17} + \dfrac{612}{17}\right) \\[5ex] \log_{3}\left(\dfrac{648}{17}\right)$ $\log_3{2x} + \log 100 \\[3ex] \log_{3}\left(2 * \dfrac{36}{17}\right) + 2 \\[5ex] \log_{3}\left(\dfrac{72}{17}\right) + \log_{3}{3^2} \\[5ex] \log_{3}\left(\dfrac{72}{17}\right) + \log_{3}{9} \\[5ex] \log_{3}\left(\dfrac{72}{17} * 9\right) \\[5ex] \log_{3}\left(\dfrac{648}{17}\right)$
(22.) WASSCE If $\log_a{y + 2} = 1 + \log_a{x}$, find x in terms of y

$1 = \log_a{a} ...Law\;4...Log \\[3ex] \log_a{y + 2} = 1 + \log_a{x} \\[3ex ] 1 + \log_a{x} = \log_a{y + 2} \\[3ex] \log_a{a} + \log_a{x} = \log_a{y + 2} \\[3ex] \log_a{ax} = \log_a{y + 2} \\[3ex] same\;\;base;\;\;cancel\;\;log;\;\;equate\;\;terms \\[3ex] ax = y + 2 \\[3ex] x = \dfrac{y + 2}{a}$
(23.) WASSCE Solve the equation: $2\log x - \log(1 - x) = \log(2 - x)$

$2\log x - \log(1 - x) = \log(2 - x) \\[3ex] \log x^2 - \log(1 - x) = \log(2 - x) \\[3ex] \log\left(\dfrac{x^2}{1 - x}\right) = \log(2 - x) \\[5ex] same\;\;base;\;\;cancel\;\;log;\;\;equate\;\;terms \\[3ex] \dfrac{x^2}{1 - x} = 2 - x \\[5ex] x^2 = (1 - x)(2 - x) \\[3ex] x^2 = 2 - x - 2x + x^2 \\[3ex] x^2 = 2 - 3x + x^2 \\[3ex] x^2 - x^2 + 3x = 2 \\[3ex] 3x = 2 \\[3ex] x = \dfrac{2}{3} \\[5ex]$ Check
$x = \dfrac{2}{3}$
LHS RHS
$2\log x - \log(1 - x) \\[5ex] 2\log \left(\dfrac{2}{3}\right) - \log\left(1 - \dfrac{2}{3}\right) \\[5ex] \log \left(\dfrac{2}{3}\right)^2 - \log\left(\dfrac{3}{3} - \dfrac{2}{3}\right) \\[5ex] \log \left(\dfrac{4}{9}\right) - \log \left(\dfrac{1}{3}\right) \\[5ex] \log \left(\dfrac{4}{9} \div \dfrac{1}{3}\right) \\[5ex] \log \left(\dfrac{4}{9} * \dfrac{3}{1}\right) \\[5ex] \log \left(\dfrac{4}{3}\right)$ $\log(2 - x) \\[3ex] \log \left(2 - \dfrac{2}{3}\right) \\[5ex] \log \left(\dfrac{6}{3} - \dfrac{2}{3}\right) \\[5ex] \log \left(\dfrac{4}{3}\right)$
(24.) NYSED Solve algebraically for the exact value of x:
$\log_8 {16} = x + 1$

$\log_8 {16} = x + 1 \\[3ex] 8^{x + 1} = 16 \\[3ex] 2^{3{x + 1}} = 2^4 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 3(x + 1) = 4 \\[3ex] x + 1 = \dfrac{4}{3} \\[5ex] x = \dfrac{4}{3} - 1 \\[5ex] x = \dfrac{4}{3} - \dfrac{3}{3} \\[5ex] x = \dfrac{1}{3} \\[5ex]$ Check
$x = \dfrac{1}{3}$
LHS RHS
$\log_8 {16} \\[3ex] change\;\;to\;\;base\;\;2 \\[3ex] \dfrac{\log_2 {16}}{\log_2 {8}} \\[5ex] \dfrac{\log_2 {2^4}}{\log_2 {2^3}} \\[5ex] \dfrac{4\log_2 {2}}{3\log_2 {2}} \\[5ex] \dfrac{4 * 1}{3 * 1} \\[5ex] \dfrac{4}{3}$ $x + 1 \\[3ex] \dfrac{1}{3} + 1 \\[5ex] \dfrac{1}{3} + \dfrac{3}{3} \\[5ex] \dfrac{4}{3}$
(25.) WASSCE Solve: $3\log_{10}2 - 2\log_{10}3 = 1 + \log_{10}{\left(\dfrac{1}{x}\right)}$

$3\log_{10}2 - 2\log_{10}3 = 1 + \log_{10}{\left(\dfrac{1}{x}\right)} \\[5ex] \log_{10}2^3 - \log_{10}3^2 = \log_{10}10 + \log_{10}{\left(\dfrac{1}{x}\right)} \\[5ex] \log_{10}\left(\dfrac{2^3}{3^2}\right) = \log_{10}\left[10\left(\dfrac{1}{x}\right)\right] \\[5ex] same\;\;base;\;\;cancel\;\;log;\;\;equate\;\;terms \\[3ex] \dfrac{2^3}{3^2} = 10\left(\dfrac{1}{x}\right) \\[5ex] \dfrac{8}{9} = \dfrac{10}{x} \\[5ex] 8x = 9(10) \\[3ex] 8x = 90 \\[3ex] x = \dfrac{90}{8} \\[5ex] x = \dfrac{45}{4} \\[5ex]$ Check
$x = \dfrac{45}{4}$
LHS RHS
$3\log_{10}2 - 2\log_{10}3 \\[5ex] \log_{10}2^3 - \log_{10}3^2 \\[5ex] \log_{10}\left(\dfrac{2^3}{3^2}\right) \\[5ex] \log_{10}\left(\dfrac{8}{9}\right)$ $1 + \log_{10}{\left(\dfrac{1}{x}\right)} \\[5ex] \dfrac{1}{x} = 1 \div x = 1 \div \dfrac{45}{4} = 1 * \dfrac{4}{45} = \dfrac{4}{45} \\[5ex] \implies \\[3ex] \log_{10}10 + \log_{10}\left(\dfrac{4}{45}\right) \\[5ex] \log_{10}\left(10 * \dfrac{4}{45}\right) \\[5ex] \log_{10}\left(\dfrac{40}{45}\right) \\[5ex] \log_{10}\left(\dfrac{8}{9}\right)$
(26.) ACT What real value of x satisfies the equation $\log_5{\left(25^2\right)} = 2x$?

$F.\;\; 2 \\[3ex] G.\;\; 4 \\[3ex] H.\;\; 8 \\[3ex] J.\;\; 25 \\[3ex] K.\;\; 125 \\[3ex]$

$\log_5{\left(25^2\right)} = 2x \\[3ex] 25^2 = 5^{2x} ...By\;\;Definition \\[3ex] 5^{2x} = 25^2 \\[3ex] 5^{2x} = 5^{2(2)} \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 2x = 2(2) \\[3ex] 2 \cdot x = 2 \cdot 2 \\[3ex] x = 2 \\[3ex]$ Check
$x = 2$
LHS RHS
$\log_5{\left(25^2\right)} \\[3ex] \log_5{\left(5^{2(2)}\right)} \\[3ex] \log_5{\left(5^4\right)} \\[3ex] 4\log_5{5}...Law\;5...Log \\[3ex] 4(1)...Law\;4...Log \\[3ex] 4$ $2x \\[3ex] 2(2) \\[3ex] 4$
(27.) NYSED Solve for p algebraically: $\log_{16}(p^2 - p + 4) - \log_{16}(2p + 11) = \dfrac{3}{4}$

$\log_{16}(p^2 - p + 4) - \log_{16}(2p + 11) = \dfrac{3}{4} \\[5ex] \log_{16}\left[\dfrac{p^2 - p + 4}{2p + 11}\right] = \dfrac{3}{4} \\[5ex] \dfrac{p^2 - p + 4}{2p + 11} = 16^{\dfrac{3}{4}} \\[5ex] \dfrac{p^2 - p + 4}{2p + 11} = (\sqrt[4]{16})^3 \\[5ex] \dfrac{p^2 - p + 4}{2p + 11} = 2^3 \\[5ex] \dfrac{p^2 - p + 4}{2p + 11} = 8 \\[3ex] p^2 - p + 4 = 8(2p + 11) \\[3ex] p^2 - p + 4 = 16p + 88 \\[3ex] p^2 - p - 16p + 4 - 88 = 0 \\[3ex] p^2 - 17p - 84 = 0 \\[3ex] (p + 4)(p - 21) = 0 \\[3ex] p + 4 = 0 \;\;\;OR\;\;\; p - 21 = 0 \\[3ex] p = -4 \;\;\;OR\;\;\; p = 21 \\[3ex]$ Check
$p = -4 \;\;\;OR\;\;\; p = 21$
LHS RHS
$p = -4 \\[3ex] \log_{16}(p^2 - p + 4) - \log_{16}(2p + 11) \\[3ex] \log_{16}[(-4)^2 - (-4) + 4] - \log_{16}[2(-4) + 11] \\[3ex] \log_{16}(16 + 4 + 4) - \log_{16}(-8 + 11) \\[3ex] \log_{16}(24) - \log_{16}(3) \\[3ex] \log_{16}\left(\dfrac{24}{3}\right) \\[5ex] \log_{16}{8} \\[3ex] To\;\;simplify;\;\;change\;\;to\;\;base\;\;2 \\[3ex] \dfrac{\log_2{8}}{\log_2{16}} \\[5ex] \dfrac{\log_2{2^3}}{\log_2{2^4}} \\[5ex] \dfrac{3\log_2{2}}{4\log_2{2}} \\[5ex] \dfrac{3 * 1}{4 * 1} \\[5ex] \dfrac{3}{4} \\[5ex]$
$p = 21 \\[3ex] \log_{16}(p^2 - p + 4) - \log_{16}(2p + 11) \\[3ex] \log_{16}(21^2 - 21 + 4) - \log_{16}[2(21) + 11] \\[3ex] \log_{16}(441 - 21 + 4) - \log_{16}(42 + 11) \\[3ex] \log_{16}(424) - \log_{16}(53) \\[3ex] \log_{16}\left(\dfrac{424}{53}\right) \\[5ex] \log_{16}{8} \\[3ex] To\;\;simplify;\;\;change\;\;to\;\;base\;\;2 \\[3ex] \dfrac{\log_2{8}}{\log_2{16}} \\[5ex] \dfrac{\log_2{2^3}}{\log_2{2^4}} \\[5ex] \dfrac{3\log_2{2}}{4\log_2{2}} \\[5ex] \dfrac{3 * 1}{4 * 1} \\[5ex] \dfrac{3}{4}$
$\dfrac{3}{4}$
(28.) WASSCE Solve for x if $\log_{10}(3x + 1) + \log{10}\left(\dfrac{1}{2}\right) - \log_{10}(2x - 5) = 0$

$\log_{10}(3x + 1) + \log{10}\left(\dfrac{1}{2}\right) - \log_{10}(2x - 5) = 0 \\[5ex] \log_{10}\left[(3x + 1) * \dfrac{1}{2} \div (2x - 5)\right] = 0 \\[5ex] \log_{10}\left[\dfrac{(3x + 1)}{1} * \dfrac{1}{2} \div \dfrac{(2x - 5)}{1}\right] = 0 \\[5ex] \log_{10}\left[\dfrac{(3x + 1)}{1} * \dfrac{1}{2} * \dfrac{1}{(2x - 5)}\right] = \log_{10}10^0 \\[5ex] \log_{10}\left[\dfrac{3x + 1}{2(2x - 5)}\right] = \log_{10}1 \\[5ex] same\;\;base;\;\;\;cancel\;\;log;\;\;\;equate\;\;terms \\[3ex] \dfrac{3x + 1}{2(2x - 5)} = 1 \\[5ex] \dfrac{3x + 1}{4x - 10} = 1 \\[5ex] 3x + 1 = 1(4x - 10) \\[3ex] 3x + 1 = 4x - 10 \\[3ex] 4x - 10 = 3x + 1 \\[3ex] 4x - 3x = 1 + 10 \\[3ex] x = 11 \\[3ex]$ Check
$x = 11$
LHS RHS
$\log_{10}(3x + 1) + \log{10}\left(\dfrac{1}{2}\right) - \log_{10}(2x - 5) \\[5ex] \log_{10}[3(11) + 1] + \log{10}\left(\dfrac{1}{2}\right) - \log_{10}[2(11) - 5] \\[5ex] \log_{10}(33 + 1) + \log{10}\left(\dfrac{1}{2}\right) - \log_{10}(22 - 5) \\[5ex] \log_{10}34 + \log{10}\left(\dfrac{1}{2}\right) - \log_{10}17 \\[5ex] \log_{10}\left(34 * \dfrac{1}{2} \div 17\right) \\[5ex] \log_{10}\left(34 * \dfrac{1}{2} * \dfrac{1}{17}\right) \\[5ex] \log_{10}1 \\[3ex] \log_{10}10^0 \\[3ex] 0\log_{10}10 \\[3ex] 0 * 1 \\[3ex] 0$ $0$
(29.) Solve these logarithmic equations.

$(a.)\;\; \log 5x = \log (2x + 9) \\[3ex] (b.)\;\; \log (-2a + 9) = \log (7 - 4a) \\[3ex] (c.)\;\; -10 + \log 3(n + 3) = -10 \\[3ex] (d.)\;\; -10 + \log_3{(n + 3)} = -10 \\[3ex]$

$(a.) \\[3ex] \log 5x = \log (2x + 9) \\[3ex] same\;\;log;\;\;equate\;\;terms \\[3ex] 5x = 2x + 9 \\[3ex] 5x - 2x = 9 \\[3ex] 3x = 9 \\[3ex] x = \dfrac{9}{3} \\[5ex] x = 3 \\[3ex]$ Check
$x = 3$
LHS RHS
$\log 5x \\[3ex] \log 5(3) \\[3ex] \log 15$ $\log (2x + 9) \\[3ex] \log [2(3) + 9] \\[3ex] \log (6 + 9) \\[3ex] \log 15$

$(b.) \\[3ex] \log (-2a + 9) = \log (7 - 4a) \\[3ex] same\;\;log;\;\;equate\;\;terms \\[3ex] -2a + 9 = 7 - 4a \\[3ex] -2a + 4a = 7 - 9 \\[3ex] 2a = -2 \\[3ex] a = -\dfrac{2}{2} \\[5ex] a = -1 \\[3ex]$ Check
$a = -1$
LHS RHS
$\log (-2a + 9) \\[3ex] \log [-2(-1) + 9] \\[3ex] \log (2 + 9) \\[3ex] \log 11$ $\log (7 - 4a) \\[3ex] \log [7 - 4(-1)] \\[3ex] \log (7 + 4) \\[3ex] \log 11$

$(c.) \\[3ex] -10 + \log [3(n + 3)] = -10 \\[3ex] \log [3(n + 3)] = -10 + 10 \\[3ex] \log (3n + 9) = 0 \\[3ex] \log (3n + 9) = \log 1...Law\;3...Log \\[3ex] same\;\;log;\;\;equate\;\;terms \\[3ex] 3n + 9 = 1 \\[3ex] 3n = 1 - 9 \\[3ex] 3n = -8 \\[3ex] n = -\dfrac{8}{3} \\[5ex]$ Check
$n = -\dfrac{8}{3}$
LHS RHS
$-10 + \log [3(n + 3)] \\[3ex] -10 + \log \left[3\left(-\dfrac{8}{3} + 3\right)\right] \\[5ex] -10 + \log \left[3\left(\dfrac{-8}{3} + \dfrac{9}{3}\right)\right] \\[5ex] -10 + \log \left[3\left(\dfrac{-8 + 9}{3}\right)\right] \\[5ex] -10 + \log \left[3\left(\dfrac{1}{3}\right)\right] \\[5ex] -10 + \log 1 \\[3ex] -10 + 0 \\[3ex] -10$ $-10$

$(d.) \\[3ex] -10 + \log_3{(n + 3)} = -10 \\[3ex] \log_3{(n + 3)} = -10 + 10 \\[3ex] \log_3{(n + 3)} = 0 \\[3ex] 3^0 = n + 3 ...By\;\;Definition \\[3ex] 1 = n + 3...Law\;3...Exp \\[3ex] n + 3 = 1 \\[3ex] n = 1 - 3 \\[3ex] n = -2 \\[3ex]$ Check
$n = -2$
LHS RHS
$-10 + \log_3{(n + 3)} \\[3ex] -10 + \log_3{(-2 + 3)} \\[3ex] -10 + \log_3{1} \\[3ex] -10 + 0 ...Law\;3...Log \\[3ex] -10$ $-10$
(30.) ACT If $\ln x = 2$, then x = ?

$A.\;\; 1 \\[3ex] B.\;\; \dfrac{2}{e} \\[5ex] C.\;\; 2e \\[3ex] D.\;\; e \\[3ex] E.\;\; e^2 \\[3ex]$

$\ln x = 2 \\[3ex] \log_e x = 2 \\[3ex] e^2 = x ...Definition \\[3ex] x = e^2$
(31.)

(32.) WASSCE Solve: $3\log_{10}2 - 2\log_{10}3 = 1 + \log_{10}\left(\dfrac{1}{x}\right)$

$3\log_{10}2 - 2\log_{10}3 = 1 + \log_{10}\left(\dfrac{1}{x}\right) \\[5ex] \log_{10}2^3 - \log_{10}3^2 = \log_{10}10 + \log_{10}\left(\dfrac{1}{x}\right) \\[5ex] \log_{10}8 - \log_{10}9 = \log_{10}10 + \log_{10}\left(\dfrac{1}{x}\right) \\[5ex] \log_{10}\left(\dfrac{8}{9}\right) = \log_{10}\left[10\left(\dfrac{1}{x}\right)\right] \\[5ex] \log_{10}\left(\dfrac{8}{9}\right) = \log_{10}\left(\dfrac{10}{x}\right) \\[5ex] same\;\;base;\;\;same\;\;log;\;\;equate\;\;terms \\[3ex] \dfrac{8}{9} = \dfrac{10}{x} \\[5ex] 8x = 9(10) \\[3ex] x = \dfrac{90}{8} \\[5ex] x = \dfrac{45}{4} \\[5ex]$ Check
$x = \dfrac{45}{4}$
LHS RHS
$3\log_{10}2 - 2\log_{10}3 \\[3ex] \log_{10}2^3 - \log_{10}3^2 \\[3ex] \log_{10}8 - \log_{10}9 \\[3ex] \log_{10}\left(\dfrac{8}{9}\right)$ $1 + \log_{10}\left(\dfrac{1}{x}\right) \\[5ex] \dfrac{1}{x} = 1 \div x \\[3ex] = 1 \div \dfrac{45}{4} = 1 * \dfrac{4}{45} = \dfrac{4}{45} \\[5ex] \implies \\[3ex] \log_{10}10 + \log_{10}\left(\dfrac{4}{45}\right) \\[5ex] \log_{10}\left(\dfrac{10 * 4}{45}\right) \\[5ex] \log_{10}\left(\dfrac{40}{45}\right) \\[5ex] \log_{10}\left(\dfrac{8}{9}\right)$
(33.) Solve and Check the logarithmic equations.

$(a.)\;\; \log_3 x = 4 \\[3ex] (b.)\;\; \log_4 (3x) = 4 \\[3ex] (c.)\;\; \log_2 (7x) = 2 \\[3ex] (d.)\;\; \log_4 (5x - 7) = 4 \\[3ex] (e.)\;\; \log_8 (x + 4) = \log_8 8 \\[3ex]$

$(a.) \\[3ex] \log_3 x = 4 \\[3ex] x = 3^4...Relationship \\[3ex] x = 81 \\[3ex]$ Check
$x = 81$
LHS RHS
$\log_3 x \\[3ex] \log_3 81 \\[3ex] \log_3 3^4 \\[3ex] 4 \log_3 3 \\[3ex] 4(1) \\[3ex] 4$ 4

$(b.) \\[3ex] \log_4 (3x) = 4 \\[3ex] 3x = 4^4...Relationship \\[3ex] 3x = 256 \\[3ex] x = \dfrac{256}{3} \\[5ex]$ Check
$x = \dfrac{256}{3}$
LHS RHS
$\log_4 (3x) \\[3ex] \log_4 \left(3 \cdot \dfrac{256}{3}\right) \\[5ex] \log_4 256 \\[3ex] \log_4 4^4 \\[3ex] 4 \log_4 4 \\[3ex] 4(1) \\[3ex] 4$ 4

$(c.) \\[3ex] \log_2 (7x) = 2 \\[3ex] 7x = 2^2...Relationship \\[3ex] 7x = 4 \\[3ex] x = \dfrac{4}{7} \\[5ex]$ Check
$x = \dfrac{256}{3}$
LHS RHS
$\log_2 (7x) \\[3ex] \log_2 \left(7 \cdot \dfrac{4}{7}\right) \\[5ex] \log_2 4 \\[3ex] \log_2 2^2 \\[3ex] 2 \log_2 2 \\[3ex] 2(1) \\[3ex] 2$ 2

$(d.) \\[3ex] \log_4 (5x - 7) = 4 \\[3ex] 5x - 7 = 4^4...Relationship \\[3ex] 5x - 7 = 256 \\[3ex] 5x = 256 + 7 \\[3ex] 5x = 263 \\[3ex] x = \dfrac{263}{5} \\[5ex]$ Check
$x = \dfrac{263}{5}$
LHS RHS
$\log_4 (5x - 7) \\[3ex] \log_4 \left(5 \cdot \dfrac{263}{5} - 7\right) \\[5ex] \log_4 (263 - 7) \\[3ex] \log_4 256 \\[3ex] \log_4 4^4 \\[3ex] 4 \log_4 4 \\[3ex] 4(1) \\[3ex] 4$ 4

$(e.) \\[3ex] \log_8 (x + 4) = \log_8 8 \\[3ex] same\;\;base;\;\;equate\;\;terms \\[3ex] x + 4 = 8 \\[3ex] x = 8 - 4 \\[3ex] x = 4 \\[3ex]$ Check
$x = 4$
LHS RHS
$\log_8 (x + 4) \\[3ex] \log_8 (4 + 4) \\[3ex] \log_8 8 \\[3ex] 1 ...Law\;4...Log$ $\log_8 8 \\[3ex] 1 ...Law\;4...Log$
(34.) ACT What is the value of the positive real number x such that $\log_x\left(\dfrac{1}{25}\right) = -2$

$F.\;\; 5 \\[3ex] G.\;\; 50 \\[3ex] H.\;\; \dfrac{1}{50} \\[5ex] J.\;\; \dfrac{1}{5} \\[5ex] K.\;\; \dfrac{25}{2} \\[5ex]$

$\log_x\left(\dfrac{1}{25}\right) = -2 \\[5ex] x^{-2} = \dfrac{1}{25} ...Definition \\[5ex] \dfrac{1}{x^2} = \dfrac{1}{25} ...Law\;6...Exp \\[5ex] 1 = 1 \\[3ex] \implies \\[3ex] x^2 = 25 \\[3ex] x = \pm \sqrt{25} \\[3ex] x = \pm 5 \\[3ex]$ However, because the base of a logarithm cannot be negative:

$x = 5 \\[3ex]$ Check
$x = 5$
LHS RHS
$\log_x\left(\dfrac{1}{25}\right) \\[5ex] \log_5\left(\dfrac{1}{25}\right) \\[5ex] \log_5 (25)^{-1} \\[3ex] \log_5 (5^2)^{-1} \\[3ex] \log_5 5^{(2 * -1)} ... Law\;5...Exp \\[3ex] \log_5 (5^{-2}) \\[3ex] -2 \log_5 5 ...Law\;5...Log \\[3ex] -2 * 1 ...Law\;4...Log \\[3ex] -2$ $-2$
(35.) Solve and Check the logarithmic equations.

$(a.)\;\; \log_9 |x| = 3 \\[3ex] (b.)\;\; \log_2 |2x - 3| = \log_2 21 \\[3ex] (c.)\;\; \dfrac{1}{2}\log_9 x = 2\log_9 2 \\[5ex] (d.)\;\; \dfrac{1}{2}\log_3 x = 3\log_3 3 \\[5ex] (e.)\;\; 6\log_2 x = -\log_2 64 \\[3ex]$

$(a.) \\[3ex] \log_9 |x| = 3 \\[3ex] |x| = 9^3 ...Relationship \\[3ex] |x| = 729 \\[3ex] x = 729 \;\;\;OR\;\;\; -x = 729 \\[3ex] x = 729 \;\;\;OR\;\;\; x = -729 \\[3ex]$ Check
$x = -729, 729$
LHS RHS
$\log_9 |x| \\[3ex] x = -729 \\[3ex] \log_9 |-729| \\[3ex] \log_9 729 \\[3ex] \log_9 9^3 \\[3ex] 3\log_9 9 \\[3ex] 3(1) \\[3ex] 3$
$\log_9 |x| \\[3ex] x = 729 \\[3ex] \log_9 729 \\[3ex] \log_9 9^3 \\[3ex] 3\log_9 9 \\[3ex] 3(1) \\[3ex] 3$
3

$(b.) \\[3ex] \log_2 |2x - 3| = \log_2 21 \\[3ex] same\;\;base;\;\;equate\;\;terms \\[3ex] |2x - 3| = 21 \\[3ex] (2x - 3) = 21 \;\;\;OR\;\;\; -(2x - 3) = 21 \\[3ex] 2x - 3 = 21 \;\;\;OR\;\;\; 2x - 3 = -21 \\[3ex] 2x = 21 + 3 \;\;\;OR\;\;\; 2x = -21 + 3 \\[3ex] 2x = 24 \;\;\;OR\;\;\; 2x = -18 \\[3ex] x = \dfrac{24}{2} \;\;\;OR\;\;\; x = -\dfrac{18}{2} \\[5ex] x = 12 \;\;\;OR\;\;\; x = -9 \\[3ex]$ Check
$x = -9, 12$
LHS RHS
$\log_2 |2x - 3| \\[3ex] x = -9 \\[3ex] \log_2 |2(-9) - 3| \\[3ex] \log_2 |-18 - 3| \\[3ex] \log_2 |-21| \\[3ex] \log_2 21$
$\log_2 |2x - 3| \\[3ex] x = 12 \\[3ex] \log_2 |2(12) - 3| \\[3ex] \log_2 |24 - 3| \\[3ex] \log_2 |21| \\[3ex] \log_2 21$
$\log_2 21$

$(c.) \\[3ex] \dfrac{1}{2}\log_9 x = 2\log_9 2 \\[5ex] \log_9 x^{\dfrac{1}{2}} = \log_9 2^2 ...Law\;5...Log \\[5ex] \log_9 \sqrt{x} = \log_9 4 \\[3ex] same\;\;base;\;\;equate\;\;terms \\[3ex] \sqrt{x} = 4 \\[3ex] x = 4^2 \\[3ex] x = 16 \\[3ex]$ Check
$x = 16$
LHS RHS
$\dfrac{1}{2}\log_9 x \\[5ex] \dfrac{1}{2}\log_9 16 \\[5ex] \log_9 16^{\dfrac{1}{2}} \\[5ex] \log_9 \sqrt{16} \\[3ex] \log_9 4 \\[3ex] \log_9 2^2 \\[3ex] 2\log_9 2$ $2\log_9 2$

$(d.) \\[3ex] \dfrac{1}{2}\log_3 x = 3\log_3 3 \\[5ex] \log_3 x^{\dfrac{1}{2}} = \log_3 3^3 ...Law\;5...Log \\[5ex] \log_3 \sqrt{x} = \log_3 27 \\[3ex] same\;\;base;\;\;equate\;\;terms \\[3ex] \sqrt{x} = 27 \\[3ex] x = 27^2 \\[3ex] x = 729 \\[3ex]$ Check
$x = 729$
LHS RHS
$\dfrac{1}{2}\log_3 x \\[5ex] \dfrac{1}{2}\log_3 729 \\[5ex] \log_3 729^{\dfrac{1}{2}} \\[5ex] \log_3 \sqrt{729} \\[3ex] \log_3 27 \\[3ex] \log_3 3^3 \\[3ex] 3\log_3 3$ $3\log_3 3$

$(e.) \\[3ex] 6\log_2 x = -\log_2 64 \\[3ex] \log_2 x^6 = \log_2 64^{-1} \\[3ex] same\;\;base;\;\;equate\;\;terms \\[3ex] x^6 = 64^{-1} x^6 = \dfrac{1}{64}...Law\;6...Exp \\[5ex] x = \sqrt[6]{\dfrac{1}{64}} \\[5ex] x = \dfrac{1}{2} \\[5ex]$ Check
$x = \dfrac{1}{2}$
LHS RHS
$6\log_2 x \\[3ex] 6\log_2 \left(\dfrac{1}{2}\right) \\[5ex] \log_2 \left(\dfrac{1}{2}\right)^6 \\[5ex] \log_2 \left(\dfrac{1}{64}\right) \\[5ex] \log_2 64^{-1} \\[3ex] -1\log_2 64$ $-\log_2 64$
(36.)

(37.) Solve and Check the logarithmic equations.

$(a.)\;\; 2\log_{12} (x - 4) + \log_{12} 4 = 2 \\[3ex] (b.)\;\; 2\log_5 (x - 2) + \log_5 5 = 3 \\[3ex] (c.)\;\; \log x + \log(x - 48) = 2 \\[3ex] (d.)\;\; \log(2x + 1) = 1 + \log(x - 4) \\[3ex] (e.)\;\; 6\log_2 x = -\log_2 64 \\[3ex]$

$(a.) \\[3ex] 2\log_{12} (x - 4) + \log_{12} 4 = 2 \\[3ex] \log_{12} (x - 4)^2 + \log_{12} 4 = 2 \\[3ex] \log_{12}[4(x - 4)^2] = 2 \\[3ex] 4(x - 4)^2 = 12^2 ...Relationship \\[3ex] 4(x - 4)^2 = 144 \\[3ex] 4[(x - 4)(x - 4)] = 144 \\[3ex] (x - 4)(x - 4) = \dfrac{144}{4} \\[5ex] x^2 - 4x - 4x + 16 = 36 \\[3ex] x^2 - 8x + 16 - 36 = 0 \\[3ex] x^2 - 8x - 20 = 0 \\[3ex] (x + 2)(x - 10) = 0 \\[3ex] x + 2 = 0 \;\;\;OR\;\;\; x - 10 = 0 \\[3ex] x = -2 \;\;\;OR\;\;\; x = 10 \\[3ex]$ Check
$x = -2, 10$
LHS RHS
$2\log_{12} (x - 4) + \log_{12} 4 \\[3ex] x = -2 \\[3ex] 2\log_{12} (-2 - 4) + \log_{12} 4 \\[3ex] 2\log_{12} (-6) + \log_{12} 4 \\[3ex]$ Because of the logarithm of a negative number
−2 is NOT a solution.
$2\log_{12} (x - 4) + \log_{12} 4 \\[3ex] x = 10 \\[3ex] 2\log_{12} (10 - 4) + \log_{12} 4 \\[3ex] 2\log_{12} 6 + \log_{12} 4 \\[3ex] \log_{12} 6^2 + \log_{12} 4 \\[3ex] \log_{12} 36 + \log_{12} 4 \\[3ex] \log_{12} (36 \cdot 4) \\[3ex] \log_{12} 144 \\[3ex] \log_{12} 12^2 \\[3ex] 2 \log_{12} 12 \\[3ex] 2(1) \\[3ex] 2$
2

$(b.) \\[3ex] 2\log_5 (x - 2) + \log_5 5 = 3 \\[3ex] \log_5 (x - 2)^2 + 1 = 3 \\[3ex] \log_5(x - 2)^2 = 3 - 1 \\[3ex] \log_5(x - 2)^2 = 2 \\[3ex] (x - 2)^2 = 5^2 ...Relationship \\[3ex] (x - 2)(x - 2) - 5^2 = 0 \\[3ex] x^2 - 2x - 2x + 4 - 25 = 0 \\[3ex] x^2 - 4x - 21 = 0 \\[3ex] (x + 3)(x - 7) = 0 \\[3ex] x + 3 = 0 \;\;\;OR\;\;\; x - 7 = 0 \\[3ex] x = -3 \;\;\;OR\;\;\; x = 7 \\[3ex]$ Check
$x = -3, 7$
LHS RHS
$2\log_5 (x - 2) + \log_5 5 \\[3ex] x = -3 \\[3ex] 2\log_5 (-3 - 2) + \log_5 5 \\[3ex] 2\log_5 (-5) + \log_5 5 \\[3ex]$ Because of the logarithm of a negative number
−3 is NOT a solution.
$2\log_5 (x - 2) + \log_5 5 \\[3ex] x = 7 \\[3ex] 2\log_5 (7 - 2) + 1 \\[3ex] 2\log_5 5 + 1 \\[3ex] 2(1) + 1 \\[3ex] 2 + 1 \\[3ex] 3$
3

$(c.) \\[3ex] \log x + \log(x - 48) = 2 \\[3ex] \log_{10}[x(x - 48)] = 2 \\[3ex] x(x - 48) = 10^2 ...Relationship \\[3ex] x^2 - 48x - 100 = 0 \\[3ex] (x + 2)(x - 50) = 0 \\[3ex] x + 2 = 0 \;\;\;OR\;\;\; x - 50 = 0 \\[3ex] x = -2 \;\;\;OR\;\;\; x = 50 \\[3ex]$ Check
$x = -2, 50$
LHS RHS
$\log x + \log(x - 48) \\[3ex] x = -2 \\[3ex] \log (-2) + \log(-2 - 48) \\[3ex]$ Because of the logarithm of a negative number
−2 is NOT a solution.
$\log x + \log(x - 48) \\[3ex] x = 50 \\[3ex] \log 50 + \log(50 - 48) \\[3ex] \log 50 + \log 2 \\[3ex] \log(50 \cdot 2) \\[3ex] \log 100 \\[3ex] 2$
2

$(d.) \\[3ex] \log(2x + 1) = 1 + \log(x - 4) \\[3ex] \log(2x + 1) = \log 10 + \log(x - 4) \\[3ex] \log(2x + 1) = \log[(10(x - 4))] \\[3ex] same\;\;base;\;\;equate\;\;terms \\[3ex] 2x + 1 = 10(x - 4) \\[3ex] 2x + 1 = 10x - 40 \\[3ex] 10x - 40 = 2x + 1 \\[3ex] 10x - 2x = 1 + 40 \\[3ex] 8x = 41 \\[3ex] x = \dfrac{41}{8} \\[5ex]$
(38.) ACT When $\log_4 x = -3$, what is x?

$A.\;\; \dfrac{1}{64} \\[5ex] B.\;\; \dfrac{1}{12} \\[5ex] C.\;\; -12 \\[3ex] D.\;\; -64 \\[3ex] E.\;\; There\;\;is\;\;no\;\;such\;\;value\;\;of\;\;x \\[3ex]$

$\log_4 x = -3 \\[3ex] 4^{-3} = x ...By\;\;Definition \\[3ex] x = 4^{-3} \\[3ex] x = \dfrac{1}{4^3} ... Law\;6...Exp \\[5ex] x = \dfrac{1}{64} \\[5ex]$ Check
$x = \dfrac{1}{64}$
LHS RHS
$\log_4 x \\[3ex] = \log_4 \left(\dfrac{1}{64}\right) \\[5ex] = \log_4 \left(\dfrac{1}{4^3}\right) \\[5ex] = \log_4 4^{-3} ...Law\;6...Exp \\[3ex] = -3 \log_4 4 ...Law\;5...Log \\[3ex] = -3 * 1 ...Law\;4...Log \\[3ex] = -3$ $-3$
(39.)

(40.)

(41.)

(42.)