Samuel Dominic Chukwuemeka (SamDom For Peace)

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

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.
Check your solutions as applicable.
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[3]{\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[3]{\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 $