If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Solved Examples on Logarithmic Equations

Samuel Dominic Chukwuemeka (SamDom For Peace) Prerequisite Topics:
(1.) Linear Equations
(2.) Quadratic Equations
(3.) Factoring
(4.) Literal Equations

Calculators: Logarithmic Equations Calculator

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.
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 did 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}$ = $10 ...Law\: 7...Log$

$ \rightarrow 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)} $




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(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)$4


$ 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.)


(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.)


(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.)


(36.)


(37.)


(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.)






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(41.)


(42.)