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

# Exponential and Logarithmic Functions Applications

### Before solving these problems, please ensure you master all the laws of exponents and logarithms.

For ACT Students
The ACT is a timed exam...$60$ minutes for $60$ questions
So, you should try to solve each question correctly and timely
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, 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.

For each problem:
specify the application
specify the discipline/subject
show all work.

(1.) ACT The number of decibels, $d$ produced by an audio source can be modeled by the equation $d = 10\log{\left(\dfrac{I}{K}\right)}$, where $I$ is the sound intensity of the audio source and $K$ is a constant.
How many decibels are produced by an audio source whose sound intensity is $1,000$ times the value of $K$?

Application: Intensity of sound/sound waves
Discipline: Physics

$d = 10\log{\left(\dfrac{I}{K}\right)} \\[5ex] I = 1000 * K = 1000K \\[3ex] d = 10\log{\left(\dfrac{1000K}{K}\right)} \\[5ex] = 10 * \log{1000} \\[3ex] = 10 * 3 \\[3ex] = 30 \:\:decibels$
(2.) $27^x = 3$

First Method - By Exponents $27^x = 3 \\[2ex] 3^{3x} = 3 \\[2ex] Base\: is\: the\: same \\[2ex] Equate\: the\: exponents \\[2ex] 3x = 3 \\[2ex] x = \dfrac{1}{3}$
Second Method - By Logarithms $27^x = 3 \\[2ex] x = \log_{27}{3} ...Relationship \\[2ex] x = \dfrac{\log_3{3}}{\log_3{27}} ...Law\: 6...Log \\[3ex] \log_3{3} = 1 ...Law\: 4...Log \\[2ex] \log_3{27} = \log_3{3^3} = 3\log_3{3} ...Law\: 5...Log \\[2ex] \log_3{27} = 3 * 1 = 3 \\[2ex] x = \dfrac{1}{3}$
Check
 LHS $27^x \\[2ex] = 27^{\dfrac{1}{3}} \\[2ex] = \sqrt[3]{27} ...Law\: 7...Exp \\[2ex] = 3$ RHS $3$
(3.) $4^{6x - 1} = 16$

First Method - By Exponents $4^{6x - 1} = 16 \\[2ex] 4^{6x - 1} = 4^2 \\[2ex] Base\: is\: the\: same \\[2ex] Equate\: the\: exponents \\[2ex] 6x - 1 = 2 \\[2ex] 6x = 2 + 1 \\[2ex] 6x = 3 \\[2ex] x = \dfrac{3}{6} \\[3ex] x = \dfrac{1}{2}$
Second Method - By Logarithms $4^{6x - 1} = 16 \\[2ex] Introduce\: \log_4\: to\: both\: sides \\[2ex] \rightarrow \log_4{4^{6x - 1}} = \log_4{16} \\[2ex] \log_4{4^{6x - 1}} = (6x - 1)\log_4{4} ...Law\: 5...Log \\[2ex] \log_4{4} = 1...Law\: 4...Log \\[2ex] \log_4{4^{6x - 1}} = (6x - 1) * 1 = 6x - 1 \\[2ex] \log_4{16} = \log_4{4^2} = 2 * \log_4{4} ...Law\:5 ...Log \\[2ex] \log_4{16} = 2 * 1 = 2 \\[2ex] \rightarrow 6x - 1 = 2 \\[2ex] 6x = 2 + 1 \\[2ex] 6x = 3 \\[2ex] x = \dfrac{3}{6} \\[3ex] x = \dfrac{1}{2}$
Check
 LHS $4^{6x - 1} \\[2ex] 4^{6 * \dfrac{1}{2} - 1} \\[2ex] 4^{3 - 1} \\[2ex] 4^{2} \\[2ex] 16$ RHS $16$
(4.) $x^7 = 128$

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

First Method - By Exponents $x^{-3} = 125 \\[2ex] Multiply\: both\: exponents\: by\: -\dfrac{1}{3} \\[3ex] x^{-3 * -\dfrac{1}{3}} = 125^{-\dfrac{1}{3}} \\[3ex] x^{-3 * -\dfrac{1}{3}} = x \\[3ex] 125^{-\dfrac{1}{3}} = \sqrt[3]{(125)^{-1}} ...Law\: 7...Exp \\[3ex] (125)^{-1} = \dfrac{1}{125^{1}} ...Law\: 6...Exp \\[3ex] \rightarrow x = \sqrt[3]{\dfrac{1}{125}} \\[3ex] x = \dfrac{1}{5}$
Second Method - By Logarithms $x^{-3} = 125 \\[2ex] Introduce\: \log_5\: to\: both\: sides \\[2ex] \rightarrow \log_5{x^{-3}} = \log_5{125} \\[2ex] \log_5{x^{-3}} = -3\log_5{x} ...Law\: 5...Log \\[2ex] \log_5{125} = \log_5{5^3} = 3\log_5{5} ... Law\: 5...Log \\[2ex] \log_5{5} = 1...Law\: 4...Log \\[2ex] \log_5{125} = 3 * 1 = 3 \\[2ex] \rightarrow -3\log_5{x} = 3 \\[2ex] Divide\: both\: sides\: by\: -3 \\[2ex] \log_5{x} = -1 \\[2ex] -1 = \log_5{5^{-1}} ...Laws\: 4\: and\: 5...Log \\[2ex] \rightarrow \log_5{x} = \log_5{5^{-1}} \\[2ex] LogBase\: is\: the\: same \\[2ex] Equate\: the\: terms \\[2ex] x = 5^{-1} \\[2ex] 5^{-1} = \dfrac{1}{5} ...Law\: 6...Exp \\[3ex] \rightarrow x = \dfrac{1}{5}$
Check
 LHS $x^{-3} \\[2ex] \left(\dfrac{1}{5}\right)^{-3} \\[3ex] = \dfrac{(1)^{-3}}{(5)^{-3}} ...Law\: 5...Exp \\[3ex] 1^{-3} = \dfrac{1}{1^3} ...Law\: 6...Exp \\[3ex] 1^{-3} = \dfrac{1}{1} = 1 \\[2ex] 5^{-3} = \dfrac{1}{5^3} ...Law\: 6...Exp \\[3ex] 5^{-3} = \dfrac{1}{125} \\[2ex] = 1 \div \dfrac{1}{125} \\[3ex] = 1 * 125 \\[2ex] = 125$ RHS $125$
(6.) $x$${\dfrac{1}{4}} = 2 First Method - By Exponents x^{\dfrac{1}{4}} = 2 \\[2ex] Multiply\: both\: exponents\: by\: 4 \\[2ex] x^{\dfrac{1}{4} * 4} = 2^4 \\[2ex] x = 16 Second Method - By Logarithms x^{\dfrac{1}{4}} = 2 \\[2ex] Introduce\: \log_2\: to\: both\: sides \\[2ex] \rightarrow \log_2{x^{\dfrac{1}{4}}} = \log_2{2} \\[2ex] \log_2{x^{\dfrac{1}{4}}} = \dfrac{1}{4}\log_2{x} ...Law\: 5...Log \\[2ex] \log_2{2} = 1...Law\: 4...Log \\[2ex] \rightarrow \dfrac{1}{4}\log_2{x} = 1 \\[2ex] Multiply\: both\: sides\: by\: 4 \\[2ex] \log_2{x} = 4 \\[2ex] 4 = \log_2{2^4} ...Laws\: 4\: and\: 5...Log \\[2ex] \rightarrow \log_2{x} = \log_2{2^4} \\[2ex] LogBase\: is\: the\: same \\[2ex] Equate\: the\: terms \\[2ex] x = 2^4 \\[2ex] x = 16 Check  LHS x^{\dfrac{1}{4}} \\[3ex] 16^{\dfrac{1}{4}} \\[3ex] \sqrt[4]{16} \\[3ex] 2 RHS 2 (7.) 9^{x^{2}} * 3^{3x} = 9 First Method - By Exponents 9^{x^{2}} * 3^{3x} = 9 \\[3ex] 9 = 3^2 \\[3ex] 9^{x^{2}} = 3^{{2} ({x^{2}})} \\[3ex] 3^{{2} ({x^{2}})} = 3^{2x^{2}} ...Law\: 5...Exp \\[3ex] \rightarrow 3^{2x^{2}} * 3^{3x} = 3^{2x^2 + 3x} ...Law\: 1...Exp \\[3ex] 3^{2x^2 + 3x} = 3^2 \\[3ex] Same\: base \\[2ex] Equate\: the\: exponents \\[2ex] 2x^2 + 3x = 2 \\[3ex] 2x^2 + 3x - 2 = 0 \\[3ex] 2x^2 + 4x - x - 2 = 0 \\[3ex] 2x(x + 2) - 1(x + 2) = 0 \\[3ex] x + 2 = 0\:\:\: OR\:\:\: 2x - 1 = 0 \\[3ex] x = -2\:\:\: OR\:\:\: 2x = 1 \\[3ex] x = -2\:\:\: OR\:\:\: x = \dfrac{1}{2} \\ Second Method - By Logarithms 9^{x^{2}} * 3^{3x} = 9 \\[3ex] Introduce\: \log_3\: to\: both\: sides \\[3ex] \rightarrow \log_3{(9^{x^{2}} * 3^{3x})} = \log_3{9} \\[3ex] \log_3{(9^{x^{2}} * 3^{3x})} = \log_3{9^{x^{2}}} + \log_3{3^{3x}} ...Law\: 1...Log \\[3ex] \log_3{9^{x^{2}}} = x^2\log_3{9} ... Law\: 5...Log \\[3ex] \log_3{9} = \log_3{3^2} = 2\log_3{3} ...Law\: 5...Log \\[3ex] \log_3{3} = 1...Law\: 4...Log \\[3ex] 2\log_3{3} = 2 * 1 = 2 \\[3ex] x^2\log_3{9} = x^2 * 2 * 1 = 2x^2 \\[3ex] \log_3{3^{x}} = 3x\log_3{3} ... Law\: 5...Log \\[3ex] 3x\log_3{3} = 3x * 1 = 3x \\[3ex] \rightarrow x^2\log_3{9} + 3x\log_3{3} = 2\log_3{3} \\[3ex] 2x^2 + 3x = 2 \\[3ex] 2x^2 + 3x - 2 = 0 \\[3ex] 2x^2 + 4x - x - 2 = 0 \\[3ex] 2x(x + 2) - 1(x + 2) = 0 \\[3ex] x + 2 = 0\:\:\: OR\:\:\: 2x - 1 = 0 \\[3ex] x = -2\:\:\: OR\:\:\: 2x = 1 \\[3ex] x = -2\:\:\: OR\:\:\: x = \dfrac{1}{2} Check  LHS 9^{x^{2}} * 3^{3x} \\[3ex] x = -2 \\[3ex] = 9^{(-2)^{2}} * 3^{3(-2)} \\[3ex] = 9^{4} * 3^{-6} \\[3ex] = 3^{2(4)} * 3^{-6} \\[3ex] 3^{2(4)} = 3^{2 * 4} = 3^8 ...Law\: 5...Exp \\[3ex] = 3^8 * 3^{-6} \\[3ex] = 3^{8 + (-6)} ...Law\: 1...Exp \\[3ex] = 3^{8 - 6} \\[3ex] = 3^2 \\[3ex] = 9 LHS 9^{x^{2}} * 3^{3x} \\[3ex] x = \dfrac{1}{2} \\[3ex] = 9^{\left(\dfrac{1}{2}\right)^{2}} * 3^{3\left(\dfrac{1}{2}\right)} \\[3ex] = 9^{\dfrac{1}{4}} * 3^{\dfrac{3}{2}} \\[3ex] = 3^{2\left(\dfrac{1}{4}\right)} * 3^{\dfrac{3}{2}} \\[3ex] 3^{2\left(\dfrac{1}{4}\right)} = 3^{2 * \dfrac{1}{4}} = 3^{\dfrac{1}{2}} ...Law\: 5...Exp \\[3ex] = 3^{\dfrac{1}{2}} * 3^{\dfrac{3}{2}} \\[3ex] = 3^{\left(\dfrac{1}{2} + \dfrac{3}{2}\right)} ...Law\: 1...Exp \\[3ex] = 3^{\left(\dfrac{1 + 3}{2}\right)} \\[3ex] = 3^{\dfrac{4}{2}} \\[3ex] = 3^2 \\[3ex] = 9 RHS 9 (8.) \left(\sqrt[5]{7}\right)$$1 - 4x$ = $7^{x^2}$

First Method - By Exponents
$\left(\sqrt[5]{7}\right)$$1 - 4x = 7^{x^2} \sqrt[5]{7} = 7$$\dfrac{1}{5}$ $...Law\: 7...Exp$

$\left(\sqrt[5]{7}\right)$$1 - 4x = 7$$\left(\dfrac{1}{5}\right) * (1 - 4x)$

$7$$\left(\dfrac{1}{5}\right) * (1 - 4x) = 7$$\left(\dfrac{1 - 4x}{5}\right)$ $...Law\: 7...Exp$

$\rightarrow 7^{\left(\dfrac{1 - 4x}{5}\right)} = 7^{x^2} \\[3ex] Same\: base \\[2ex] Equate\: the\: exponents \\[2ex] \dfrac{1 - 4x}{5} = x^2 \\[3ex] 1 - 4x = 5x^2 \\[3ex] 0 = 5x^2 + 4x - 1 \\[3ex] 5x^2 + 4x - 1 = 0 \\[3ex] 5x^2 + 5x - x - 1 = 0 \\[3ex] 5x(x + 1) - 1(x + 1) = 0 \\[3ex] x + 1 = 0\:\:\: OR\:\:\: 5x - 1 = 0 \\[3ex] x = -1\:\:\: OR\:\:\: 5x = 1 \\[3ex] x = -1\:\:\: OR\:\:\: x = \dfrac{1}{5} \\$
Second Method - By Logarithms
$\left(\sqrt[5]{7}\right)$$1 - 4x = 7^{x^2} \sqrt[5]{7} = 7$$\dfrac{1}{5}$ $...Law\: 7...Exp$

$\left(\sqrt[5]{7}\right)$$1 - 4x = 7$$\left(\dfrac{1}{5}\right) * (1 - 4x)$

$7$$\left(\dfrac{1}{5}\right) * (1 - 4x) = 7$$\left(\dfrac{1 - 4x}{5}\right)$ $...Law\: 7...Exp$

$\rightarrow 7^{\left(\dfrac{1 - 4x}{5}\right)} = 7^{x^2} \\[3ex] Introduce\: \log_7\: to\: both\: sides \\[3ex] \rightarrow \log_7{7^{\left(\dfrac{1 - 4x}{5}\right)}} = \log_7{7^{x^2}} \\[3ex] \log_7{7^{\left(\dfrac{1 - 4x}{5}\right)}} = \left(\dfrac{1 - 4x}{5}\right) \log_7{7} ... Law\: 5...Log \\[3ex] \log_7{7^{x^2}} = x^2 \log_7{7} ... Law\: 5...Log \\[3ex] \log_7{7} = 1...Law\: 4...Log \\[3ex] \rightarrow \dfrac{1 - 4x}{5} = x^2 \\[3ex] 1 - 4x = 5x^2 \\[3ex] 0 = 5x^2 + 4x - 1 \\[3ex] 5x^2 + 4x - 1 = 0 \\[3ex] 5x^2 + 5x - x - 1 = 0 \\[3ex] 5x(x + 1) - 1(x + 1) = 0 \\[3ex] x + 1 = 0\:\:\: OR\:\:\: 5x - 1 = 0 \\[3ex] x = -1\:\:\: OR\:\:\: 5x = 1 \\[3ex] x = -1\:\:\: OR\:\:\: x = \dfrac{1}{5}$
Check
 LHS $\left(\sqrt[5]{7}\right)$$1 - 4x x = -1 1 - 4x = 1 - 4(-1) = 1 + 4 = 5 \left(\sqrt[5]{7}\right)$$5$ $\left(\sqrt[5]{7}\right) = 7^{\dfrac{1}{5}} ...Law\: 7...Exp \\[3ex] \left(\sqrt[5]{7}\right)^5 = 7^{\left(\dfrac{1}{5}\right)5} \\[3ex] 7^{\left(\dfrac{1}{5}\right)5} = 7^{\left(\dfrac{1}{5} * 5\right)} ...Law\: 5...Exp \\[3ex] = 7^1 \\[3ex] = 7$ LHS $\left(\sqrt[5]{7}\right)$$1 - 4x x = \dfrac{1}{5} 1 - 4x = 1 - 4 * \dfrac{1}{5} = 1 - \dfrac{4}{5} = \dfrac{1}{5} = \left(\sqrt[5]{7}\right)$$\dfrac{1}{5}$ $\left(\sqrt[5]{7}\right) = 7^{\dfrac{1}{5}} ...Law\: 7...Exp \\[3ex] = \left(\sqrt[5]{7}\right)^5 = 7^{\left(\dfrac{1}{5}\right)\dfrac{1}{5}} \\[3ex] 7^{\left(\dfrac{1}{5}\right)\dfrac{1}{5}} = 7^{\left(\dfrac{1}{5} * \dfrac{1}{5}\right)} ...Law\: 5...Exp \\[3ex] = 7^{\dfrac{1}{25}}$ RHS $7^{x^2} \\[3ex] x = -1 \\[3ex] x^2 = (-1)^2 = 1 \\[3ex] = 7^1 \\[3ex] = 7$ RHS $7^{x^2} \\[3ex] x = \dfrac{1}{5} \\[3ex] x^2 = \left(\dfrac{1}{5}\right)^2 \\[3ex] \left(\dfrac{1}{5}\right)^2 = \dfrac{1}{25} \\[3ex] = 7^{\dfrac{1}{25}}$
(9.) $\dfrac{\sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}}}{e^{2x} \div e^{-2x}} = e^{13}$

$\dfrac{\sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}}}{e^{2x} \div e^{-2x}} = e^{13} \\[5ex] LCD = e^{2x} \div e^{-2x} \\[5ex] Multiply\: both\: sides\: by\: \left(e^{2x} \div e^{-2x}\right) \\[5ex] \left(e^{2x} \div e^{-2x}\right) * \dfrac{\sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}}}{e^{2x} \div e^{-2x}} = \left(e^{2x} \div e^{-2x}\right) * e^{13} \\[5ex] = \sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}} = \left(e^{2x} \div e^{-2x}\right) * e^{13} \\[5ex] \left(e^{2x} \div e^{-2x}\right) = \dfrac{e^{2x}}{e^{-2x}} \\[5ex] \dfrac{e^{2x}}{e^{-2x}} = e^{2x - (-2x)} ...Law\: 2...Exp \\[5ex] e^{2x - (-2x)} = e^{2x + 2x} = e^{4x} \\[5ex] \therefore \left(e^{2x} \div e^{-2x}\right) = e^{4x} \\[5ex] \sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}} = \left[\left(e^{2x} * e^{-5x}\right)^{-4}\right]^{\dfrac{1}{2}} ...Law\: 7...Exp \\[5ex] \left[\left(e^{2x} * e^{-5x}\right)^{-4}\right]^{\dfrac{1}{2}} = \left[\left(e^{2x} * e^{-5x}\right)^{-4 * \dfrac{1}{2}}\right] ...Law\: 5...Exp \\[5ex] \left[\left(e^{2x} * e^{-5x}\right)^{-4 * \dfrac{1}{2}}\right] = \left[\left(e^{2x} * e^{-5x}\right)^{-2}\right] \\[5ex] \left(e^{2x} * e^{-5x}\right) = e^{2x + (-5x)} ...Law\: 1...Exp \\[5ex] e^{2x + (-5x)} = e^{2x - 5x} = e^{-3x} \\[5ex] \left[\left(e^{2x} * e^{-5x}\right)^{-2}\right] = \left(e^{-3x}\right)^{-2} \\[5ex] \left(e^{-3x}\right)^{-2} = \left(e^{-3x * -2}\right) ...Law\: 5...Exp \\[5ex] \left(e^{-3x * -2}\right) = e^{6x} \\[5ex] \therefore \sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}} = e^{6x} \\[5ex] \rightarrow e^{6x} = e^{4x} * e^{13} \\[5ex] Divide\: both\: sides\: by\: e^{4x} \\[5ex] e^{6x} \div e^{4x} = e^{13} \\[5ex] e^{6x} \div e^{4x} = e^{6x - 4x} = e^{2x} ...Law\: 2...Exp \\[5ex] \rightarrow e^{2x} = e^{13} \\[5ex] Same\: base \\[3ex] Equate\: the\: exponents \\[3ex] 2x = 13 \\[3ex] x = \dfrac{13}{2}$
Check
 LHS $\dfrac{\sqrt{\left(e^{2x} * e^{-5x}\right)^{-4}}}{e^{2x} \div e^{-2x}} \\[5ex] x = \dfrac{13}{2} \\[5ex] = \dfrac{\sqrt{\left(e^{\left(2 * \dfrac{13}{2}\right)} * e^{\left(-5 * \dfrac{13}{2}\right)}\right)^{-4}}}{e^{\left(2 * \dfrac{13}{2}\right)} \div e^{\left(-2 * \dfrac{13}{2}\right)}} \\[9ex] = \dfrac{\sqrt{\left(e^{13} * e^{\dfrac{-65}{2}}\right)^{-4}}}{e^{13} \div e^{-13}} \\[9ex] e^{13} * e^{\dfrac{-65}{2}} = e^{\left(13 + \dfrac{-65}{2}\right)}...Law\: 1...Exp \\[5ex] e^{13 + \dfrac{-65}{2}} = e^{\left(13 - \dfrac{65}{2}\right)} \\[5ex] e^{\left(13 - \dfrac{65}{2}\right)} = e^{\left(\dfrac{26}{2} - \dfrac{65}{2}\right)} \\[5ex] e^{\left(\dfrac{26}{2} - \dfrac{65}{2}\right)} = e^{\left(\dfrac{26 - 65}{2}\right)} = e^{\dfrac{-39}{2}} \\[5ex] \left(e^{\dfrac{-39}{2}}\right)^{-4} = e^{\left(\dfrac{-39}{2} * -4\right)} ...Law\: 5...Exp \\[5ex] e^{\left(\dfrac{-39}{2} * -4\right)} = e^{78} \\[5ex] \sqrt{e^{78}} = \left(e^{78}\right)^{\dfrac{1}{2}} ...Law\: 7...Exp \\[5ex] \left(e^{78}\right)^{\dfrac{1}{2}} = \left(e^{78}\right)^{\dfrac{1}{2}} \\[5ex] \left(e^{78}\right)^{\dfrac{1}{2}} = e^{\left(78 * \dfrac{1}{2}\right)} ...Law\: 5...Exp \\[5ex] e^{\left(78 * \dfrac{1}{2}\right)} = e^{39} \\[5ex] Numerator = e^{39} \\[5ex] e^{13} \div e^{-13} = e^{\left(13 - (-13)\right)} ...Law\: 2...Exp \\[5ex] e^{\left(13 - (-13)\right)} = e^{\left(13 + 13\right)} = e^{26} \\[5ex] Denominator = e^{26} \\[5ex] = \dfrac{e^{39}}{e^{26}} \\[5ex] = e^{\left(39 - 26\right)} ...Law\: 2...Exp \\[5ex] = e^{13}$ RHS $e^{13}$
(10.) $4^{x - 4} = 64(3^x)$

$4^{x - 4} = 64(3^x)$

$3$ is not a multiple of $4$

In that regard, we shall not be solving it By Exponents.

We shall solve it By Logarithms.

$Introduce\: \log_4\: to\: both\: sides \\[3ex] \log_4{4^{x - 4}} = \log_4{[64(3^x)]} \\[3ex] \log_4{4^{x - 4}} = (x - 4)\log_4{4} ...Law\: 1...Log \\[3ex] \log_4{4} = 1 ...Law\: 4...Log \\[3ex] (x - 4)\log_4{4} = (x - 4) * 1 = (x - 4) \\[3ex] \log_4{[64(3^x)]} = \log_4{64} + \log_4{3^x} \\[3ex] \log_4{64} = \log_4{4^3} = 3\log_4{4} ...Law\: 5...Log \\[3ex] \log_4{64} = 3 * 1 = 3 \\[3ex] \log_4{3^x} = x\log_4{3} \\[3ex] \rightarrow x - 4 = 3 + x\log_4{3} \\[3ex] x - x\log_4{3} = 3 + 4 \\[3ex] x(1 - \log_4{3}) = 7 \\[3ex] x = \dfrac{7}{1 - \log_4{3}}$

OR

$\log_4{3} = \dfrac{\log4}{\log3} ...Law\: 6...Log \\[3ex] \log_4{3} = 0.792481250 \\[3ex] x = \dfrac{7}{1 - \log_4{3}} \\[3ex] x = \dfrac{7}{1 - 0.792481250} \\[3ex] x = \dfrac{7}{0.207518750} \\[3ex] x = 33.73189176$
Check
 LHS $4^{x - 4} \\[3ex] x = 33.73189176 \\[3ex] 4^{33.73189176 - 4} \\[3ex] 4^{29.73189176} \\[3ex] 7.950281244 * 10^{17}$ RHS $64(3^x) \\[3ex] 64(3^{33.73189176}) \\[3ex] 64 * 1.242231451 * 10^{16} \\[3ex] 7.950281287 * 10^{17}$
(11.) $10^{-x} = 6^{3x}$

$10^{-x} = 6^{3x} \\[3ex] Introduce\: \log\: to\: both\: sides \\[2ex] \rightarrow \log 10^{-x} = \log 6^{3x} \\[3ex] \log 10^{-x} = -x \log 10 ...Law\: 5...Log \\[3ex] \log 6^{3x} = 3x \log 6 ...Law\: 5...Log \\[3ex] \log 10 = 1...Law\: 4...Log \\[3ex] -x \log 10 = -x * 1 = -x \\[3ex] \rightarrow -x = 3x \log 6 \\[3ex] \dfrac{-x}{3x} = \log 6 \\[3ex] \dfrac{-1}{3} = \log 6$
Technically, there is no solution.
However, what if $x = 0$?
Let us check.

Check
 LHS $x = 0 \\[2ex] 10^{-x} \\[2ex] 10^{-0} \\[2ex] 10^0 \\[2ex] 1$ RHS $x = 0 \\[2ex] 6^{3x} \\[2ex] 6^{3 * 0} \\[2ex] 6^0 \\[2ex] 1$

$\therefore x = 0$
(12.) $x^7 = 128$

First Method - By Exponents $x^7 = 128 \\[2ex] x^7 = 2^7 \\[2ex] Exponents\: are\: the\: same \\[2ex] Equate\: the\: bases \\[2ex] x = 2$
Second Method - By Logarithms $x^7 = 128 \\[2ex] Introduce\: \log_2\: to\: both\: sides \\[2ex] \log_2{x^7} = \log_2{128} \\[2ex] \log_2{x^7} = 7\log_2{x} ...Law\: 5...Log \\[2ex] \log_2{128} = \log_2{2^7} = 7\log_2{2} ... Law\: 5...Log \\[2ex] \log_2{2} = 1...Law\: 4...Log \\[2ex] \log_2{128} = 7 * 1 = 7 \\[2ex] \rightarrow 7\log_2{x} = 7 \\[2ex] Divide\: both\: sides\: by\: 7 \\[2ex] \log_2{x} = 1 \\[2ex] 1 = \log_2{2} ...Law\: 4...Log \\[2ex] \log_2{x} = \log_2{2} \\[2ex] LogBase\: is\: the\: same \\[2ex] Equate\: the\: terms \\[2ex] x = 2$
Check
 LHS $x^7 \\[2ex] 2^{7} \\[2ex] 128$ RHS $128$
(13.) WASSCE Find the value of $x$ in the expression: $\dfrac{3^{(2x + 1)}}{3^{(3x - 4)} * 3^{(6 - 7x)}} = 27^x$

$\dfrac{3^{(2x + 1)}}{3^{(3x - 4)} * 3^{(6 - 7x)}} = 27^x \\[5ex] LHS \\[3ex] \dfrac{3^{(2x + 1)}}{3^{(3x - 4)} * 3^{(6 - 7x)}} \\[3ex] 3^{(3x - 4)} * 3^{(6 - 7x)} = 3^{(3x - 4) + (6 - 7x)} ...Law\:\: 1...Exp \\[3ex] 3^{(3x - 4)} * 3^{(6 - 7x)} = 3^{3x - 4 + 6 - 7x} \\[3ex] 3^{(3x - 4)} * 3^{(6 - 7x)} = 3^{-4x + 2} \\[3ex] = \dfrac{3^{2x + 1}}{3^{-4x + 2}} = 3^{(2x + 1) - (-4x + 2)} ...Law\:\: 2...Exp \\[3ex] = 3^{2x + 1 + 4x - 2} \\[3ex] = 3^{6x - 1} \\[3ex] RHS \\[3ex] 27^x = (3^3)^x = 3^{3 * x} = 3^{3x} ...Law\:\: 5...Exp \\[3ex] LHS = RHS \\[3ex] 3^{6x - 1} = 3^{3x} \\[3ex] Same\:\: base \\[3ex] Equate\:\: exponents \\[3ex] 6x - 1 = 3x \\[3ex] 6x - 3x = 1 \\[3ex] 3x = 1 \\[3ex] x = \dfrac{1}{3} \\[5ex]$ Check
 $\underline{LHS} \\[3ex] \dfrac{3^{(2x + 1)}}{3^{(3x - 4)} * 3^{(6 - 7x)}} \\[5ex] x = \dfrac{1}{3} \\[5ex] 2x + 1 = 2 * \dfrac{1}{3} + 1 \\[5ex] 2x + 1 = \dfrac{2}{3} + \dfrac{3}{3} \\[5ex] 2x + 1 = \dfrac{2 + 3}{3} \\[5ex] 2x + 1 = \dfrac{5}{3} \\[5ex] 3x - 4 = 3 * \dfrac{1}{3} - 4 \\[5ex] 3x - 4 = 1 - 4 = -3 \\[3ex] 6 - 7x = 6 - 7 * \dfrac{1}{3} \\[5ex] 6 - 7x = 6 - \dfrac{7}{3} \\[5ex] 6 - 7x = \dfrac{18}{3} - \dfrac{7}{3} \\[5ex] 6 - 7x = \dfrac{18 - 7}{3} \\[5ex] 6 - 7x = \dfrac{11}{3} \\[5ex] = \dfrac{3^{\dfrac{5}{3}}}{3^{-3} * 3^{\dfrac{11}{3}}} \\[7ex] = \dfrac{3^{\dfrac{5}{3}}}{3^{-3 + \dfrac{11}{3}}} ...Law\:\: 1...Exp \\[7ex] = \dfrac{3^{\dfrac{5}{3}}}{3^{-\dfrac{9}{3} + \dfrac{11}{3}}} \\[7ex] = \dfrac{3^{\dfrac{5}{3}}}{3^{\dfrac{-9 + 11}{3}}} \\[7ex] = \dfrac{3^{\dfrac{5}{3}}}{3^{\dfrac{2}{3}}} \\[7ex] = 3^{\dfrac{5}{3} - \dfrac{2}{3}} ...Law\:\: 2...Exp \\[5ex] = 3^{\dfrac{5 - 2}{3}} \\[5ex] = 3^{\dfrac{3}{3}} \\[5ex] = 3^1 \\[3ex] = 3$ $\underline{RHS} \\[3ex] 27^x \\[3ex] x = \dfrac{1}{3} \\[5ex] = 27^{\dfrac{1}{3}} \\[5ex] = (3^3)^{\dfrac{1}{3}} \\[5ex] = 3^{3 * \dfrac{1}{3}} \\[5ex] = 3^1 \\[3ex] = 3$
(14.) ACT For what value of $x$ is the equation $2^{2x + 7} = 2^{15}$ true?

$2^{2x + 7} = 2^{15} \\[3ex] Same\:\: base \\[3ex] Equate\:\: exponents \\[3ex] 2x + 7 = 15 \\[3ex] 2x = 15 - 7 \\[3ex] 2x = 8 \\[3ex] x = \dfrac{8}{2} \\[5ex] x = 4 \\[3ex]$ Check
 $\underline{LHS} \\[3ex] 2^{2x + 7} \\[3ex] x = 4 \\[3ex] 2^{2(4) + 7} \\[3ex] 2^{8 + 7} \\[3ex] 2^{15}$ $\underline{RHS} \\[3ex] 2^{15}$