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# Exponential and Logarithmic Functions Applications

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

For JAMB, NZQA, and CMAT 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 WASCCE is a question for the WASCCE 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.

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.) Atmospheric pressure $P$ in pounds per square inch is represented by the formula $P = 14.7e^{-0.21x}$, where $x$ is the number of miles above sea level.
To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of $8.781$ pounds per square inch?
(Hint: there are $5,280$ feet in a mile)

$P = 14.7e^{-0.21x} \\[3ex] 14.7e^{-0.21x} = P \\[3ex] P = 8.781 \\[3ex] 14.7e^{-0.21x} = 8.781 \\[3ex] e^{-0.21x} = \dfrac{8.781}{14.7} \\[5ex] e^{-0.21x} = 0.5973469388 \\[3ex] Take\;\;the\;\;natural\;\;logarithm\;\;of\;\;both\;\;sides \\[3ex] \ln e^{-0.21x} = \ln 0.5973469388 \\[3ex] -0.21x = -0.5152571974 \\[3ex] x = \dfrac{-0.5152571974}{-0.21} \\[5ex] x = 2.453605702\;\;miles \\[3ex]$ Let $k$ be the height of the mountain in feet.
Proportional Reasoning
mile feet
$1$ $5280$
$2.453605702$ $k$

$\dfrac{5280}{1} = \dfrac{k}{2.453605702} \\[5ex] k * 1 = 5280(2.453605702) \\[3ex] k = 12955.03811 \\[3ex] k \approx 12955\;\;feet$
(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.) ACT Hikers' World Foods sells raisin-nut mix in bulk to stores.
The dollar amount per pound, P(x), for a store to purchase x pounds of raisin-nut mix from Hikers' World is given by the function below.
$P(x) = 3.50 + 0.90^x$
To the nearest $\$0.01$, which of the following dollar values is equal to the total price for a store to purchase 100 pounds of raison-nut mix in a single order from Hikers' World?$ A.\;\; \$350.00 \\[3ex] B.\;\; \$359.00 \\[3ex] C.\;\; \$440.00 \\[3ex] D.\;\; \$616.00 \\[3ex] E.\;\; \$903.50 \\[3ex]$

$P(x) = 3.50 + 0.90^x \\[3ex] x = 100 \\[3ex] P(100) = 3.50 + 0.90^{100} \\[3ex] P(100) = 3.5 + 0.00002656139889 \\[3ex] P(100) = 3.500026561 \;\;per\;\;pound \\[3ex] For\;\;100\;\;pounds: \\[3ex] P(100) = 3.500026561(100) \\[3ex] P(100) = 350.0026561 \\[3ex] P(100) \approx \$350.00 $(5.)$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{(125)^{-1}} ...Law\: 7...Exp \\[3ex] (125)^{-1} = \dfrac{1}{125^{1}} ...Law\: 6...Exp \\[3ex] \rightarrow x = \sqrt{\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{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{7}\right)$$1 - 4x$=$7^{x^2}$First Method - By Exponents$\left(\sqrt{7}\right)$$1 - 4x = 7^{x^2} \sqrt{7} = 7$$\dfrac{1}{5}...Law\: 7...Exp\left(\sqrt{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{7}\right)$$1 - 4x = 7^{x^2} \sqrt{7} = 7$$\dfrac{1}{5}...Law\: 7...Exp\left(\sqrt{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{7}\right)$$1 - 4x x = -1 1 - 4x = 1 - 4(-1) = 1 + 4 = 5 \left(\sqrt{7}\right)$$5 \left(\sqrt{7}\right) = 7^{\dfrac{1}{5}} ...Law\: 7...Exp \\[3ex] \left(\sqrt{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{7}\right)$$1 - 4x x = \dfrac{1}{5} 1 - 4x = 1 - 4 * \dfrac{1}{5} = 1 - \dfrac{4}{5} = \dfrac{1}{5} = \left(\sqrt{7}\right)$$\dfrac{1}{5} \left(\sqrt{7}\right) = 7^{\dfrac{1}{5}} ...Law\: 7...Exp \\[3ex] = \left(\sqrt{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 $This is a contradiction. Technically, there is no solution. However, what if$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} \$