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# 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 any wrong answer.

Solve each logarithmic equation.
Use at least two methods as necessary.
Show all Work.

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

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

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

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

First Method - By Exponents $\log_x{\dfrac{1}{25}} = -2 \\[3ex] x^{-2} = \dfrac{1}{25} ...Relationship \\[2ex] x^{-2} = 25^{-1}...Law\: 6...Exp \\[3ex] Multiply\: both\: exponents\: by\: -\dfrac{1}{2} \\[5ex] x^{\left(-2 * -\dfrac{1}{2}\right)} = 25^{\left(-1 * -\dfrac{1}{2}\right)}...Law\: 5...Exp \\[5ex] x = 25^{\dfrac{1}{2}} \\[5ex] 25^{\dfrac{1}{2}} = \sqrt{25} ...Law\: 7...Exp \\[5ex] \rightarrow x = \sqrt{25} \\[3ex] x = 5$
Second Method - By Logarithms $\log_x{\dfrac{1}{25}} = -2 \\[3ex] -2 = \log_x{x^{-2}} ...Laws\: 4\: and\: 5...Log \\[5ex] \rightarrow \log_x{\dfrac{1}{25}} = \log_x{x^{-2}} \\[5ex] LogBase\: is\: the\: same \\[2ex] Equate\: the\: terms \\[5ex] \dfrac{1}{25} = x^{-2} \\[5ex] x^{-2} = \dfrac{1}{25} \\[5ex] x^{-2} = \dfrac{1}{5^2} \\[5ex] \dfrac{1}{5^2} = 5^{-2}...Law\: 6...Exp \\[5ex] \rightarrow x^{-2} = 5^{-2} \\[5ex] Exponents\: are\: the\: same \\[3ex] Equate\: the\: base \\[3ex] x = 5$
Check
 LHS $\log_x{\dfrac{1}{25}} \\[5ex] = \log_5{\dfrac{1}{25}} \\[5ex] \dfrac{1}{25} = 25^{-1} ...Law\: 6...Exp \\[5ex] 25^{-1} = 5^2{-1} \\[5ex] 5^2{-1} = 5^{(2 * -1)} = 5^{-2} ...Law\: 5...Exp \\[5ex] = \log_5{5^{-2}} \\[5ex] = -2\log_5{5} ...Law\: 5...Log \\[5ex] \log_5{5} = 1...Law\: 4...Log \\[2ex] = -2 * 1 \\[2ex] = -2$ 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.) $\ln (p + 12) + \ln (p - 5) = 2 \ln p$

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

$\log_4{(p + 27)} - \log_4{(p + 7)} = \log_4{p} \\[3ex] \log_4{(p + 27)} - \log_4{(p + 7)} = \log_4{\left(\dfrac{p + 27}{p + 7}\right)} ...Law\: 2...Log \\[5ex] \rightarrow \log_4{\left(\dfrac{p + 27}{p + 7}\right)} = \log_4{p} \\[5ex] Same\: base \\[2ex] Equate\: the\: terms \\[2ex] \dfrac{p + 27}{p + 7} = p \\[3ex] LCD = p + 7 \\[3ex] Multiply\: both\: sides\: by\: the\: LCD \\[3ex] p + 27 = p(p + 7) \\[3ex] p + 27 = p^2 + 7p \\[3ex] 0 = p^2 + 7p - p - 27 \\[3ex] 0 = p^2 + 6p - 27 \\[3ex] p^2 + 6p - 27 = 0 \\[3ex] (p + 9)(p - 3) = 0 \\[3ex] p + 9 = 0\:\:\: OR\:\:\: p - 3 = 0 ...Zero\: Product\: Property \\[3ex] p = -9\:\:\: OR\:\:\: p = 3 \\[3ex] Log\: of\: a\: negative\: number\: DNE ...based\: on\: the\: RHS \\[3ex] \therefore p = 3$
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 $\log_4{p} \\[3ex] p = 3 \\[3ex] = \log_4{3}$
(11.) ACT When $\log_5{x} = -2$, what is $x$?

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

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

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

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

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