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 Exponential Equations

Samuel Dominic Chukwuemeka (SamDom For Peace) An exponential equation is an equation containing exponents and/or exponential functions.
We solve exponential equations using the Laws of Exponents.

Prerequisite Topics:
(1.) Linear Equations
(2.) Quadratic Equations
(3.) Factoring
(4.) Literal Equations

Calculators: Exponential 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 and NZQA 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 exponential 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.) $3^x = 27$


First Method: By Exponents

$ 3^x = 27 \\[3ex] 3^x = 3^3 \\[3ex] Base\: is\: the\: same \\[3ex] Equate\: the\: exponents \\[3ex] x = 3 \\[3ex] $ Second Method: By Logarithms

$ 3^x = 27 \\[3ex] x = \log_3{27} ...Relationship \\[3ex] x = \log_3{3^3} = 3\log_3{3} ...Law\: 5...Log \\[3ex] \log_3{3} = 1 ...Law\: 4...Log \\[3ex] x = 3 * 1 \\[3ex] x = 3 $
Check

LHS
$ 3^x \\[3ex] 3^3 \\[3ex] 27 $

RHS
$ 27 $

(2.) $27^x = 3$


First Method: By Exponents

$ 27^x = 3 \\[3ex] 3^{3x} = 3 \\[3ex] Base\: is\: the\: same \\[3ex] Equate\: the\: exponents \\[3ex] 3x = 3 \\[3ex] x = \dfrac{1}{3} \\[5ex] $ Second Method: By Logarithms

$ 27^x = 3 \\[3ex] x = \log_{27}{3} ...Relationship \\[3ex] x = \dfrac{\log_3{3}}{\log_3{27}} ...Law\: 6...Log \\[3ex] \log_3{3} = 1 ...Law\: 4...Log \\[3ex] \log_3{27} = \log_3{3^3} = 3\log_3{3} ...Law\: 5...Log \\[3ex] \log_3{27} = 3 * 1 = 3 \\[3ex] x = \dfrac{1}{3} $
Check

LHS
$ 27^x \\[3ex] = 27^{\dfrac{1}{3}} \\[3ex] = \sqrt[3]{27} ...Law\: 7...Exp \\[3ex] = 3 $

RHS
$ 3 $

(3.) $4^{6x - 1} = 16$


First Method: By Exponents

$ 4^{6x - 1} = 16 \\[3ex] 4^{6x - 1} = 4^2 \\[3ex] Base\:\; is\:\; the\:\; same \\[3ex] Equate\:\; the\:\; exponents \\[3ex] 6x - 1 = 2 \\[3ex] 6x = 2 + 1 \\[3ex] 6x = 3 \\[3ex] x = \dfrac{3}{6} \\[3ex] x = \dfrac{1}{2} \\[5ex] $ Second Method: By Logarithms

$ 4^{6x - 1} = 16 \\[3ex] Introduce\: \log_4\: to\: both\: sides \\[3ex] \rightarrow \log_4{4^{6x - 1}} = \log_4{16} \\[3ex] \log_4{4^{6x - 1}} = (6x - 1)\log_4{4} ...Law\: 5...Log \\[3ex] \log_4{4} = 1...Law\: 4...Log \\[3ex] \log_4{4^{6x - 1}} = (6x - 1) * 1 = 6x - 1 \\[3ex] \log_4{16} = \log_4{4^2} = 2 * \log_4{4} ...Law\:5 ...Log \\[3ex] \log_4{16} = 2 * 1 = 2 \\[3ex] \rightarrow 6x - 1 = 2 \\[3ex] 6x = 2 + 1 \\[3ex] 6x = 3 \\[3ex] x = \dfrac{3}{6} \\[3ex] x = \dfrac{1}{2} \\[5ex] $ Check

LHS
$ 4^{6x - 1} \\[3ex] 4^{6 * \dfrac{1}{2} - 1} \\[3ex] 4^{3 - 1} \\[3ex] 4^{2} \\[3ex] 16 $

RHS
$ 16 $

(4.) $x^7 = 128$


First Method: By Exponents

$ x^7 = 128 \\[3ex] x^7 = 2^7 \\[3ex] Exponents\: are\: the\: same \\[3ex] Equate\: the\: bases \\[3ex] x = 2 $
Second Method: By Logarithms

$ x^7 = 128 \\[3ex] Introduce\: \log_2\: to\: both\: sides \\[3ex] \log_2{x^7} = \log_2{128} \\[3ex] \log_2{x^7} = 7\log_2{x} ...Law\: 5...Log \\[3ex] \log_2{128} = \log_2{2^7} = 7\log_2{2} ... Law\: 5...Log \\[3ex] \log_2{2} = 1...Law\: 4...Log \\[3ex] \log_2{128} = 7 * 1 = 7 \\[3ex] \rightarrow 7\log_2{x} = 7 \\[3ex] Divide\: both\: sides\: by\: 7 \\[3ex] \log_2{x} = 1 \\[3ex] 1 = \log_2{2} ...Law\: 4...Log \\[3ex] \log_2{x} = \log_2{2} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] x = 2 $
Check

LHS
$ x^7 \\[3ex] 2^{7} \\[3ex] 128 $

RHS
$ 128 $

(5.) $x^{-3} = 125$


First Method: By Exponents

$ x^{-3} = 125 \\[3ex] 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 \\[3ex] Introduce\: \log_5\: to\: both\: sides \\[3ex] \rightarrow \log_5{x^{-3}} = \log_5{125} \\[3ex] \log_5{x^{-3}} = -3\log_5{x} ...Law\: 5...Log \\[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] \rightarrow -3\log_5{x} = 3 \\[3ex] Divide\: both\: sides\: by\: -3 \\[3ex] \log_5{x} = -1 \\[3ex] -1 = \log_5{5^{-1}} ...Laws\: 4\: and\: 5...Log \\[3ex] \rightarrow \log_5{x} = \log_5{5^{-1}} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] x = 5^{-1} \\[3ex] 5^{-1} = \dfrac{1}{5} ...Law\: 6...Exp \\[3ex] \rightarrow x = \dfrac{1}{5} $
Check

LHS
$ x^{-3} \\[3ex] \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 \\[3ex] 5^{-3} = \dfrac{1}{5^3} ...Law\: 6...Exp \\[3ex] 5^{-3} = \dfrac{1}{125} \\[3ex] = 1 \div \dfrac{1}{125} \\[3ex] = 1 * 125 \\[3ex] = 125 $

RHS
$ 125 $

(6.) $x$${\dfrac{1}{4}}$ = $2$


First Method: By Exponents

$ x^{\dfrac{1}{4}} = 2 \\[3ex] Multiply\: both\: exponents\: by\: 4 \\[3ex] x^{\dfrac{1}{4} * 4} = 2^4 \\[3ex] x = 16 $
Second Method: By Logarithms

$ x^{\dfrac{1}{4}} = 2 \\[3ex] Introduce\: \log_2\: to\: both\: sides \\[3ex] \rightarrow \log_2{x^{\dfrac{1}{4}}} = \log_2{2} \\[3ex] \log_2{x^{\dfrac{1}{4}}} = \dfrac{1}{4}\log_2{x} ...Law\: 5...Log \\[3ex] \log_2{2} = 1...Law\: 4...Log \\[3ex] \rightarrow \dfrac{1}{4}\log_2{x} = 1 \\[3ex] Multiply\: both\: sides\: by\: 4 \\[3ex] \log_2{x} = 4 \\[3ex] 4 = \log_2{2^4} ...Laws\: 4\: and\: 5...Log \\[3ex] \rightarrow \log_2{x} = \log_2{2^4} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] x = 2^4 \\[3ex] x = 16 \\[3ex] $ 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 \\[3ex] Equate\: the\: exponents \\[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} \\ $
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 \\[3ex] Equate\: the\: exponents \\[3ex] \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 \\[3ex] \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 \\[3ex] 10^{-x} \\[3ex] 10^{-0} \\[3ex] 10^0 \\[3ex] 1 $

RHS
$ x = 0 \\[3ex] 6^{3x} \\[3ex] 6^{3 * 0} \\[3ex] 6^0 \\[3ex] 1 $


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


First Method: By Exponents

$ x^7 = 128 \\[3ex] x^7 = 2^7 \\[3ex] Exponents\: are\: the\: same \\[3ex] Equate\: the\: bases \\[3ex] x = 2 $
Second Method: By Logarithms

$ x^7 = 128 \\[3ex] Introduce\: \log_2\: to\: both\: sides \\[3ex] \log_2{x^7} = \log_2{128} \\[3ex] \log_2{x^7} = 7\log_2{x} ...Law\: 5...Log \\[3ex] \log_2{128} = \log_2{2^7} = 7\log_2{2} ... Law\: 5...Log \\[3ex] \log_2{2} = 1...Law\: 4...Log \\[3ex] \log_2{128} = 7 * 1 = 7 \\[3ex] \rightarrow 7\log_2{x} = 7 \\[3ex] Divide\: both\: sides\: by\: 7 \\[3ex] \log_2{x} = 1 \\[3ex] 1 = \log_2{2} ...Law\: 4...Log \\[3ex] \log_2{x} = \log_2{2} \\[3ex] LogBase\: is\: the\: same \\[3ex] Equate\: the\: terms \\[3ex] x = 2 $
Check

LHS
$ x^7 \\[3ex] 2^{7} \\[3ex] 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} $

(15.) WASSCE Given that $2^m * \left(\dfrac{1}{8}\right)^n = 128$ and

$4^m \div 2^{-4n} = \dfrac{1}{16}$, find the value of $m - n$


$ 2^m * \left(\dfrac{1}{8}\right)^n = 128 \\[5ex] 2^m * \left(\dfrac{1}{2^3}\right)^n = 128 \\[5ex] 2^m * \left(2^{-3}\right)^n = 2^7 \\[3ex] 2^m * 2^{-3n} = 2^7 \\[3ex] 2^{m + -3n} = 2^7 \\[3ex] 2^{m - 3n} = 2^7 \\[3ex] Same\;\;base \\[3ex] Equate\;\;the\;\;exponents \\[3ex] m - 3n = 7...eqn.(1) \\[3ex] 4^m \div 2^{-4n} = \dfrac{1}{16} \\[5ex] {2^2}^m \div 2^{-4n} = 16^{-1} \\[3ex] 2^{2m} \div 2^{-4n} = {2^4}^{-1} \\[3ex] 2^{2m - -4n} = 2^{-4} \\[3ex] 2^{2m + 4n} = 2^{-4} \\[3ex] Same\;\;base \\[3ex] Equate\;\;the\;\;exponents \\[3ex] 2m + 4n = -4 \\[3ex] 2(m + 2n) = 2(-2) \\[3ex] m + 2n = -2...eqn.(2) \\[3ex] Eliminate\;\;m \rightarrow eqn.(2) - eqn.(1) \\[3ex] 2n - (-3n) = -2 - 7 \\[3ex] 2n + 3n = -9 \\[3ex] 5n = -9 \\[3ex] n = -\dfrac{9}{5} \\[5ex] From\;\;eqn.(1) \\[3ex] m = 7 + 3n \\[3ex] m = 7 + 3\left(-\dfrac{9}{5}\right) \\[5ex] m = 7 + -\dfrac{27}{5} \\[5ex] m = \dfrac{35}{5} - \dfrac{27}{5} \\[5ex] m = \dfrac{35 - 27}{5} \\[5ex] m = \dfrac{8}{5} \\[5ex] m - n \\[3ex] = \dfrac{8}{5} - -\dfrac{9}{5} \\[5ex] = \dfrac{8}{5} + \dfrac{9}{5} \\[5ex] = \dfrac{8 + 9}{5} \\[5ex] = \dfrac{17}{5} $
(16.) NSC Solve for $x:$
$2^{x + 2} + 2^x = 20$


$ 2^{x + 2} + 2^x = 20 \\[3ex] 2^x * 2^2 + 2^x = 20 \\[3ex] 2^x * 4 + 2^x = 20 \\[3ex] 2^x(4 + 1) = 20 \\[3ex] 2^x * 5 = 20 \\[3ex] 2^x = \dfrac{20}{5} \\[5ex] 2^x = 4 \\[3ex] 2^x = 2^2 \\[3ex] x = 2 \\[3ex] $ Check
$2^{x + 2} + 2^x = 20$
$x = 2$

$ \underline{LHS} \\[3ex] 2^{x + 2} + 2^x \\[3ex] 2^{2 + 2} + 2^2 \\[3ex] 2^4 + 2^2 \\[3ex] 16 + 4 \\[3ex] 20 $

$ \underline{RHS} \\[3ex] 20 $

(17.) WASSCE If $2^x + 2^{(x - 1)} = 48$, find the value of x


$ 2^x + 2^{(x - 1)} = 48 \\[3ex] 2^x + \dfrac{2^x}{2^1} = 48 \\[5ex] 2 * 2^x + 2^x = 2(48) \\[3ex] 2^x(2 + 1) = 96 \\[3ex] 2^x * 3 = 96 \\[3ex] 2^x = \dfrac{96}{3} \\[5ex] 2^x = 32 \\[3ex] 2^x = 2^5 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] x = 5 \\[3ex] $ Check
$x = 5$
LHS RHS
$ 2^x + 2^{(x - 1)} \\[3ex] 2^5 + 2^{5 - 1} \\[3ex] 32 + 2^4 \\[3ex] 32 + 16 \\[3ex] 48 $ $48$
(18.) NSC Solve for x: $2 * 3^x = 81 - 3^x$


$ 2 * 3^x = 81 - 3^x \\[3ex] 2 * 3^x + 3^x = 81 \\[3ex] 3^x(2 + 1) = 81 \\[3ex] 3^x * 3 = 81 \\[3ex] 3^x = \dfrac{81}{3} \\[5ex] 3^x = 27 \\[3ex] 3^x = 3^3 \\[3ex] x = 3 \\[3ex] $ Check
$2 * 3^x = 81 - 3^x$
$x = 3$

$ \underline{LHS} \\[3ex] 2 * 3^x \\[3ex] 2 * 3^3 \\[3ex] 2 * 27 \\[3ex] 54 $

$ \underline{RHS} \\[3ex] 81 - 3^x \\[3ex] 81 - 3^3 \\[3ex] 81 - 27 \\[3ex] 54 $

(19.) Determine the exact solution for $e^{2x} - 6e^x - 72 = 0$


$ e^{2x} - 6e^x - 72 = 0 \\[3ex] e^{x(2)} - 6e^x - 72 = 0 \\[3ex] Let\;\;e^x = p \\[3ex] p^2 - 6p - 72 = 0 \\[3ex] (p - 12)(p + 6) = 0 \\[3ex] p - 12 = 0 \;\;OR\;\; p + 6 = 0 \\[3ex] p = 12 \;\;OR\;\; p = -6 \\[3ex] But: \\[3ex] e^x = p \\[3ex] When\;\; p = 12 \\[3ex] e^x = 12 \\[3ex] Take\;\;the\;\;natural\;\;logarithm\;\;of\;\;both\;\;sides \\[3ex] \ln e^x = \ln 12 \\[3ex] x = \ln 12 \\[3ex] When\;\; p = -6 \\[3ex] e^x = -6 \\[3ex] Take\;\;the\;\;natural\;\;logarithm\;\;of\;\;both\;\;sides \\[3ex] \ln e^x = \ln (-6) \\[3ex] Argument\;\;cannot\;\;be\;\;negative \implies No\;\;solution \\[3ex] \therefore x = 12 \\[3ex] $ Check
$e^{2x} - 6e^x - 72 = 0$
$x = \ln 12$
LHS RHS
$ e^{2x} - 6e^x - 72 \\[3ex] e^{2 * \ln 12} - 6e^{\ln 12} - 72 \\[3ex] e^{\ln 12^2} - 6 * 12 - 72 \\[3ex] 12^2 - 72 - 72 \\[3ex] 144 - 72 - 72 \\[3ex] 0 $ $0$
(20.) NZQA Solve the equation $2^x * 2^{3x - 8} = 16$


$ 2^x * 2^{3x - 8} = 16 \\[3ex] 2^x * 2^{3x - 8} = 2^4 \\[3ex] 2^{x + (3x - 8)} = 2^4 ...Law\;1...Exp \\[3ex] 2^{x + 3x - 8} = 2^4 \\[3ex] 2^{4x - 8} = 2^4 \\[3ex] Same\;\;base;\;\;Equate\;\;the\;\;exponents \\[3ex] 4x - 8 = 4 \\[3ex] 4x = 4 + 8 \\[3ex] 4x = 12 \\[3ex] x = \dfrac{12}{4} \\[5ex] x = 3 \\[3ex] $ Check
$x = 3$
LHS RHS
$ 2^x * 2^{3x - 8} \\[3ex] 2^3 * 2^{3(3) - 8} \\[3ex] 8 * 2^{9 - 8} \\[3ex] 8 * 2^1 \\[3ex] 8 * 2 \\[3ex] 16 $ 16




Top




(21.) KCSE Solve the equation $8^{x + 1} - 2^{3x - 1} = 120$


$ 8^{x + 1} - 2^{3x - 1} = 120 \\[3ex] 2^{3(x + 1)} - 2^{3x - 1} = 120 \\[3ex] 2^{3x + 3} - 2^{3x - 1} = 120 \\[3ex] 2^{3x} * 2^3 - \left(2^{3x} \div 2^{1}\right) = 120 \\[3ex] 2^{x(3)} * 8 - \dfrac{2^{x(3)}}{2^{1}} = 120 \\[5ex] 8 * 2^{x(3)} - \dfrac{2^{x(3)}}{2} = 120 \\[5ex] Let\;\;2^x = p \\[3ex] \implies \\[3ex] 8 * p^3 - \dfrac{p^3}{2} = 120 \\[5ex] LCD = 2 \\[3ex] 2 * 8 * p^3 - p^3 = 2(120) \\[3ex] 16p^3 - p^3 = 240 \\[3ex] 15p^3 = 240 \\[3ex] p^3 = \dfrac{240}{15} \\[5ex] p^3 = 16 \\[3ex] \implies \\[3ex] \left(2^x\right)^3 = 16 \\[3ex] 2^{3x} = 2^4 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 3x = 4 \\[3ex] x = \dfrac{4}{3} \\[5ex] $ Check
$x = \dfrac{4}{3}$
LHS RHS
$ 8^{x + 1} - 2^{3x - 1} \\[3ex] 8^{\dfrac{4}{3} + 1} - 2^{3\left(\dfrac{4}{3}\right) - 1} \\[5ex] 8^{\dfrac{4}{3} + \dfrac{3}{3}} - 2^{4 - 1} \\[5ex] 8^{\dfrac{7}{3}} - 2^{3} \\[5ex] (\sqrt[3]{8})^7 - 8 \\[3ex] 2^7 - 8 \\[3ex] 128 - 8 \\[3ex] 120 $ $120$
(22.) MEHA Solve $2^{x + 1} = 4$


$ 2^{x + 1} = 4 \\[3ex] 2^{x + 1} = 2^2 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] x + 1 = 2 \\[3ex] x = 2 - 1 \\[3ex] x = 1 \\[3ex] $ Check
$x = 1$
$ 2^{x + 1} \\[3ex] 2^{1 + 1} \\[3ex] 2^2 \\[3ex] 4 $ $4$
(23.) NSC Solve for x: $\dfrac{5^{2x} - 1}{5^x + 1} = 4$


$ \underline{First\;\;Method:\;\;Substitution} \\[3ex] \dfrac{5^{2x} - 1}{5^x + 1} = 4 \\[5ex] \dfrac{5^{x(2)} - 1}{5^x + 1} = 4 \\[5ex] Let\;\; p = 5^x \\[3ex] \implies \\[3ex] \dfrac{p^2 - 1}{p + 1} = 4 \\[5ex] p^2 - 1 = 4(p + 1) \\[3ex] p^2 - 1 = 4p + 4 \\[3ex] p^2 - 1 - 4p - 4 = 0 \\[3ex] p^2 - 4p - 5 = 0 \\[3ex] (p + 1)(p - 5) = 0 \\[3ex] p + 1 = 0 \;\;\;OR\;\;\; p - 5 = 0 \\[3ex] p = -1 \;\;\;OR\;\;\; p = 5 \\[3ex] Recall:\;\; p = 5^x \\[3ex] 5^x = p \\[3ex] when\;\; p = -1 \\[3ex] 5^x = -1 \\[3ex] Introduce\;\;\log\;\;to\;\;both\;\;sides \\[3ex] \log 5^x = \log (-1) \\[3ex] x \log 5 = DNE...log\;\;of\;\;a\;\;negative\;\;number\;\;DNE \\[3ex] when\;\;p = 5 \\[3ex] 5^x = 5^1 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] x = 1 \\[3ex] \therefore x = 1 \\[5ex] \underline{Second\;\;Method:\;\;Difference\;\;of\;\;Two\;\;Squares} \\[3ex] \dfrac{5^{2x} - 1}{5^x + 1} = 4 \\[5ex] \dfrac{5^{x(2)} - 1^2}{5^x + 1} = 4 \\[5ex] \dfrac{(5^x + 1)(5^x - 1)}{5^x + 1} = 4 \\[5ex] 5^x - 1 = 4 \\[3ex] 5^x = 4 + 1 \\[3ex] 5^x = 5 \\[3ex] 5^x = 5^1 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] x = 1 \\[3ex] $ Check
$x = 1$
LHS RHS
$ \dfrac{5^{2x} - 1}{5^x + 1} \\[5ex] \dfrac{5^{2(1)} - 1}{5^1 + 1} \\[5ex] \dfrac{5^{2} - 1}{5 + 1} \\[5ex] \dfrac{25 - 1}{6} \\[5ex] \dfrac{24}{6} \\[5ex] 4 $ $4$
(24.) NZQA Solve the equation $2^{2y} * 2^{2y - 12} = 16$


$ 2^{2y} * 2^{2y - 12} = 16 \\[3ex] 2^{2y} * 2^{2y - 12} = 2^4 \\[3ex] 2^{2y + (2y - 12)} = 2^4 ...Law\;1...Exp \\[3ex] 2^{2y + 2y - 12} = 2^4 \\[3ex] 2^{4y - 12} = 2^4 \\[3ex] Same\;\;base;\;\;Equate\;\;the\;\;exponents \\[3ex] 4y - 12 = 4 \\[3ex] 4y = 4 + 12 \\[3ex] 4y = 16 \\[3ex] y = \dfrac{16}{4} \\[5ex] y = 4 \\[3ex] $ Check
$y = 4$
LHS RHS
$ 2^{2y} * 2^{2y - 12} \\[3ex] 2^{2(4)} * 2^{2(4) - 12} \\[3ex] 2^{8} * 2^{8 - 12} \\[3ex] 2^8 * 2^{-4} \\[3ex] 2^{8 + -4}...Law\;1...Exp \\[3ex] 2^{8 - 4} \\[3ex] 2^4 \\[3ex] 16 $ $16$
(25.)


(26.) NZQA For what values of x will $5 * 5^{3x} = 5^{-2x^2}$?


$ 5 * 5^{3x} = 5^{-2x^2} \\[3ex] 5^1 * 5^{3x} = 5^{-2x^2} \\[3ex] 5^{1 + 3x} = 5^{-2x^2} \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 1 + 3x = -2x^2 \\[3ex] 2x^2 + 3x + 1 = 0 \\[3ex] 2x^2 + 2x + x + 1 = 0 \\[3ex] 2x(x + 1) + 1(x + 1) = 0 \\[3ex] (x + 1)(2x + 1) = 0 \\[3ex] x + 1 = 0 \;\;\;OR\;\;\; 2x + 1 = 0 \\[3ex] x = -1 \;\;\;OR\;\;\; 2x = -1 \\[3ex] x = -1 \;\;\;OR\;\;\; x = -\dfrac{1}{2} \\[5ex] $ Check
$x = -1 \;\;\;OR\;\;\; x = -\dfrac{1}{2}$
LHS RHS
$ x = -1 \\[3ex] 5 * 5^{3x} \\[3ex] 5 * 5^{3(-1)} \\[3ex] 5^1 * 5^{-3} \\[3ex] 5^{1 + -3} \\[3ex] 5^{1 - 3} \\[3ex] 5^{-2} \\[3ex] $
$ x = -\dfrac{1}{2} \\[5ex] 5 * 5^{3x} \\[3ex] 5 * 5^{3\left(-\dfrac{1}{2}\right)} \\[5ex] 5^1 * 5^{\left(-\dfrac{3}{2}\right)} \\[5ex] 5^{1 + -\dfrac{3}{2}} \\[5ex] 5^{\dfrac{2}{2} - \dfrac{3}{2}} \\[5ex] 5^{\dfrac{2 - 3}{2}} \\[5ex] 5^{-\dfrac{1}{2}} $
$ x = -1 \\[3ex] 5^{-2x^2} \\[3ex] 5^{-2(1)^2} \\[3ex] 5^{-2(1)} \\[3ex] 5^{-2} \\[3ex] $
$ x = -\dfrac{1}{2} \\[5ex] 5^{-2x^2} \\[3ex] 5^{-2\left(-\dfrac{1}{2}\right)^2} \\[5ex] 5^{-2\left(\dfrac{1}{4}\right)} \\[5ex] 5^{-\dfrac{1}{2}} $
(27.)


(28.) NZQA Find the value of x if $3^{x + 1} = 81$


$ 3^{x + 1} = 81 \\[3ex] 3^{x + 1} = 3^4 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] x + 1 = 4 \\[3ex] x = 4 - 1 \\[3ex] x = 3 \\[3ex] $ Check
$x = 3$
LHS RHS
$ 3^{x + 1} \\[3ex] 3^{3 + 1} \\[3ex] 3^4 \\[3ex] 81 $ $81$
(29.)


(30.) NZQA Solve the equation $8^x * 4^{x^2 - 6} = 4$


$ 8^x * 4^{x^2 - 6} = 4 \\[3ex] 2^{3(x)} * 2^{2(x^2 - 6)} = 2^2 \\[3ex] 2^{3x} * 2^{2x^2 - 12} = 2^2 \\[3ex] 2^{3x + (2x^2 - 12)} = 2^2 ...Law\;1...Exp \\[3ex] 2^{3x + 2x^2 - 12} = 2^2 \\[3ex] 2^{2x^2 + 3x - 12} = 2^2 \\[3ex] Same\;\;base;\;\;Equate\;\;the\;\;exponents \\[3ex] 2x^2 + 3x - 12 = 2 \\[3ex] 2x^2 + 3x - 12 - 2 = 0 \\[3ex] 2x^2 + 3x - 14 = 0 \\[3ex] 2x^2 + 7x - 4x - 14 = 0 \\[3ex] x(2x + 7) - 2(2x + 7) = 0 \\[3ex] 2x + 7 = 0 \;\;\;OR\;\;\; x - 2 = 0 \\[3ex] 2x = -7 \;\;\;OR\;\;\; x = 2 \\[3ex] x = -\dfrac{7}{2} \;\;\;OR\;\;\; x = 2 \\[5ex] $ Check
$x = 2$ OR $x = -\dfrac{7}{2}$
LHS RHS
$ x = 2 \\[3ex] 8^x * 4^{x^2 - 6} \\[3ex] 8^2 * 4^{2^2 - 6} \\[3ex] 64 * 4^{4 - 6} \\[3ex] 4^3 * 4^{-2} \\[3ex] 4^{3 + -2}...Law\;1...Exp \\[3ex] 4^{3 - 2} \\[3ex] 4^1 \\[3ex] 4 $ 4
$ x = -\dfrac{7}{2} \\[5ex] 8^x * 4^{x^2 - 6} \\[3ex] 8^{-\dfrac{7}{2}} * 4^{\left(-\dfrac{7}{2}\right)^2 - 6} \\[5ex] 2^{3\left(-\dfrac{7}{2}\right)} * 4^{\dfrac{49}{4} - \dfrac{24}{4}} \\[5ex] 2^{-\dfrac{21}{2}} * 4^{\dfrac{25}{4}} \\[5ex] 2^{-\dfrac{21}{2}} * 2^{2\left(\dfrac{25}{4}\right)} \\[5ex] 2^{-\dfrac{21}{2} + \dfrac{25}{2}} \\[5ex] 2^{\dfrac{4}{2}} \\[5ex] 2^2 \\[3ex] 4 $ 4
(31.)


(32.) NZQA Find the values of y if $25 * 5^{2y + 13} = 5^{y^2}$


$ 25 * 5^{2y + 13} = 5^{y^2} \\[3ex] 5^2 * 5^{2y + 13} = 5^{y^2} \\[3ex] 5^{2 + (2y + 13)} = 5^{x^2} \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 2 + 2y + 13 = y^2 \\[3ex] 2y + 15 = y^2 \\[3ex] y^2 = 2y + 15 \\[3ex] y^2 - 2y - 15 = 0 \\[3ex] (y + 3)(y - 5) = 0 \\[3ex] y + 3 = 0 \;\;\;OR\;\;\; y - 5 = 0 \\[3ex] y = -3 \;\;\;OR\;\;\; y = 5 \\[3ex] $ Check
$y = -3 \;\;\;OR\;\;\; y = 5$
LHS RHS
$ y = -3 \\[3ex] 25 * 5^{2y + 13} \\[3ex] 25 * 5^{2(-3) + 13} \\[3ex] 25 * 5^{-6 + 13} \\[3ex] 25 * 5^{7} \\[3ex] 5^2 * 5^7 \\[3ex] 5^{2 + 7} \\[3ex] 5^9 \\[3ex] $
$ y = 5 \\[3ex] 25 * 5^{2y + 13} \\[3ex] 25 * 5^{2(5) + 13} \\[3ex] 25 * 5^{10 + 13} \\[3ex] 25 * 5^{23} \\[3ex] 5^2 * 5^{23} \\[3ex] 5^{2 + 23} \\[3ex] 5^{25} $
$ y = -3 \\[3ex] 5^{y^2} \\[3ex] 5^{(-3)^2} \\[3ex] 5^9 \\[3ex] $
$ y = 5 \\[3ex] 5^{y^2} \\[3ex] 5^{5^2} \\[3ex] 5^{25} $
(33.)


(34.) NZQA For what values of x will $7 * 7^{1 - x} = 7^{3x^2}$?


$ 7 * 7^{1 - x} = 7^{3x^2} \\[3ex] 7^1 * 7^{1 - x} = 7^{3x^2} \\[3ex] 7^{1 + (1 - x)} = 7^{3x^2} \\[3ex] 7^{1 + 1 - x} = 7^{3x^2} \\[3ex] 7^{2 - x} = 7^{3x^2} \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 2 - x = 3x^2 \\[3ex] 0 = 3x^2 + x - 2 \\[3ex] 3x^2 + x - 2 = 0 \\[3ex] 3x^2 + 3x - 2x - 2 = 0 \\[3ex] 3x(x + 1) - 2(x + 1) = 0 \\[3ex] (x + 1)(3x - 2) = 0 \\[3ex] x + 1 = 0 \;\;\;OR\;\;\; 3x - 2 = 0 \\[3ex] x = -1 \;\;\;OR\;\;\; 3x = 2 \\[3ex] x = -1 \;\;\;OR\;\;\; x = \dfrac{2}{3} \\[5ex] $ Check
$x = -1 \;\;\;OR\;\;\; x = \dfrac{2}{3}$
LHS RHS
$ x = -1 \\[3ex] 7 * 7^{1 - x} \\[3ex] 7 * 7^{1 - -1} \\[3ex] 7^1 * 7^{1 + 1} \\[3ex] 7^1 * 7^2 \\[3ex] 7^{1 + 2} \\[3ex] 7^3 \\[3ex] $
$ x = \dfrac{2}{3} \\[5ex] 7 * 7^{1 - x} \\[3ex] 7 * 7^{1 - \dfrac{2}{3}} \\[5ex] 7 * 7^{\dfrac{3}{3} - \dfrac{2}{3}} \\[5ex] 7 * 7^{\dfrac{3 - 2}{3}} \\[5ex] 7 * 7^{\dfrac{1}{3}} \\[5ex] 7^1 * 7^{\dfrac{1}{3}} \\[5ex] 7^{1 + \dfrac{1}{3}} \\[5ex] 7^{\dfrac{3}{3} + \dfrac{1}{3}} \\[5ex] 7^{\dfrac{3 + 1}{3}} \\[5ex] 7^{\dfrac{4}{3}} $
$ x = -1 \\[3ex] 7^{3x^2} \\[3ex] 7^{3(-1)^2} \\[3ex] 7^{3(1)} \\[3ex] 7^{3} \\[3ex] $
$ x = \dfrac{2}{3} \\[5ex] 7^{3x^2} \\[3ex] 7^{3\left(\dfrac{2}{3}\right)^2} \\[5ex] 7^{3\left(\dfrac{4}{9}\right)} \\[5ex] 7^{\dfrac{4}{3}} $
(35.)


(36.) ACT All the values in the equation beow are exact.
What value of c makes the equation true? $$ (4.25 * 10^{2c + 4})(6 * 10^7) = 255 $$ $ F.\;\; -7 \\[3ex] G.\;\; -6.5 \\[3ex] H.\;\; -5 \\[3ex] J.\;\; -4.5 \\[3ex] K.\;\; -4 \\[3ex] $

$ (4.25 * 10^{2c + 4})(6 * 10^7) = 255 \\[3ex] 4.25 * 10^{2c + 4} * 6 * 10^7 = 255 \\[3ex] 4.25 * 6 * 10^{2c + 4} * 10^7 = 255 \\[3ex] 25.5 * 10^{2c + 4 + 7} = 255 \\[3ex] 10^{2c + 11} = \dfrac{255}{25.5} \\[5ex] 10^{2c + 11} = \dfrac{2550}{255} \\[5ex] 10^{2c + 11} = 10^1 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 2c + 11 = 1 \\[3ex] 2c = 1 - 11 \\[3ex] 2c = -10 \\[3ex] c = \dfrac{-10}{2} \\[5ex] c = -5 \\[3ex] $ Check
$c = -5$
LHS RHS
$ (4.25 * 10^{2c + 4})(6 * 10^7) \\[3ex] 4.25 * 10^{2(-5) + 4} * 6 * 10^7 \\[3ex] 4.25 * 6 * 10^{-10 + 4} * 10^7 \\[3ex] 25.5 * 10^{-6} * 10^7 \\[3ex] 25.5 * 10^{-6 + 7} \\[3ex] 25.5 * 10^1 \\[3ex] 255 $ $255$
(37.)


(38.) NZQA Find the value of x if $2^{x - 1} = 64$


$ 2^{x - 1} = 64 \\[3ex] 2^{x - 1} = 2^6 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] x - 1 = 6 \\[3ex] x = 6 + 1 \\[3ex] x = 7 \\[3ex] $ Check
$x = 7$
LHS RHS
$ 2^{x - 1} \\[3ex] 2^{7 - 1} \\[3ex] 2^6 \\[3ex] 64 $ $64$
(39.) ACT Which of the following is the solution set of $27^{n^2} = 9^{5n - 4}$

$ A.\;\; \left\{-4, \dfrac{2}{3}\right\} \\[5ex] B.\;\; \left\{-1, \dfrac{8}{3}\right\} \\[5ex] C.\;\; \left\{-\dfrac{2}{3}, 4\right\} \\[5ex] D.\;\; \{1, 4\} \\[5ex] E.\;\; \left\{\dfrac{4}{3}, 2\right\} \\[5ex] $

$ 27^{n^2} = 9^{5n - 4} \\[3ex] 3^{3^\left(n^2\right)} = 3^{2(5n - 4)} \\[3ex] 3^{3n^2} = 3^{10n - 8} \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 3n^2 = 10n - 8 \\[3ex] 3n^2 - 10n + 8 = 0 \\[3ex] 3n^2 - 6n - 4n + 8 = 0 \\[3ex] 3n(n - 2) - 4(n - 2) = 0 \\[3ex] (n - 2)(3n - 4) = 0 \\[3ex] n - 2 = 0 \;\;\;OR\;\;\; 3n - 4 = 0 \\[3ex] n = 2 \;\;\;OR\;\;\; 3n = 4 \\[3ex] n = 2 \;\;\;OR\;\;\; n = \dfrac{4}{3} \\[5ex] $ Check
$n = 2 \;\;\;OR\;\;\; n = \dfrac{4}{3}$
LHS RHS
$ 27^{n^2} \\[3ex] n = 2 \\[3ex] 27^{2^2} \\[3ex] 27^{2^2} \\[3ex] 27^{4} \\[3ex] 3^{3(4)} \\[3ex] 3^{12} $
$ 27^{n^2} \\[3ex] n = \dfrac{4}{3} \\[5ex] 27^{\left(\dfrac{4}{3}\right)^2} \\[5ex] 27^{\dfrac{16}{9}} \\[5ex] 3^{3 * \dfrac{16}{9}} \\[5ex] 3^{\dfrac{16}{3}} $
$ n = 2 \\[3ex] 9^{5n - 4} \\[3ex] 9^{5(2) - 4} \\[3ex] 9^{10 - 4} \\[3ex] 9^{6} \\[3ex] 3^{2(6)} \\[3ex] 3^{12} $
$ 9^{5n - 4} \\[3ex] n = \dfrac{4}{3} \\[5ex] 9^{5 * \dfrac{4}{3} - 4} \\[5ex] 9^{\dfrac{20}{3} - \dfrac{12}{3}} \\[5ex] 9^{\dfrac{20 - 12}{3}} \\[5ex] 9^{\dfrac{8}{3}} \\[5ex] 3^{2 * \dfrac{8}{3}} \\[5ex] 3^{\dfrac{16}{3}} $
(40.) NSC Given: $2^x + 2^{x + 2} = -5y + 20$

(40.1) Express $2^x$ in terms of $y$

(40.2) How many solutions for $x$ will the equation have if $y = -4$?

(40.3) Solve for $x$ if $y$ is the largest possible integer value for which $2^x + 2^{x + 2} = -5y + 20$ will have solutions.


$ (40.1) \\[3ex] 2^x + 2^{x + 2} = -5y + 20 \\[3ex] 2^x + (2^x * 2^2) = -5y + 20 \\[3ex] 2^x + (2^x * 4) = -5y + 20 \\[3ex] 2^x + 2^x * 4 = -5y + 20 \\[3ex] 2^x(1 + 4) = -5y + 20 \\[3ex] 2^x(5) = -5y + 20 \\[3ex] 2^x = \dfrac{-5y + 20}{5} \\[5ex] 2^x = \dfrac{5(-y + 4)}{5} \\[5ex] 2^x = -y + 4 \\[3ex] 2^x = 4 - y \\[3ex] (40.2) \\[3ex] y = -4 \\[3ex] 2^x = 4 - (-4) \\[3ex] 2^x = 4 + 4 \\[3ex] 2^x = 8 \\[3ex] 2^x = 2^3 \\[3ex] x = 3...only\;\;one\;\;solution \\[3ex] (40.3) \\[3ex] For:\;\; 2^x + 2^{x + 2} = -5y + 20\;\;to\;\;have\;\;solutions: \\[3ex] 2^x = 4 - y \\[3ex] 2^x \gt 0 \\[3ex] \implies 4 - y \gt 0 \\[3ex] 4 - 0 \gt y \\[3ex] 4 \gt y \\[3ex] y \lt 4 \\[3ex] Largest\;\;possible\;\;integer\;\;for\;\;y = 3 \\[3ex] Solving\;\;for\;\;x \\[3ex] 2^x = 4 - 3 \\[3ex] 2^x = 1 \\[3ex] 2^x = 2^0 \\[3ex] x = 0 $


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(41.)


(42.) NZQA Find the values of x if $36 * 6^{2x + 6} = 6^{x^2}$


$ 36 * 6^{2x + 6} = 6^{x^2} \\[3ex] 6^2 * 6^{2x + 6} = 6^{x^2} \\[3ex] 6^{2 + (2x + 6)} = 6^{x^2} \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 2 + 2x + 6 = x^2 \\[3ex] 2x + 8 = x^2 \\[3ex] x^2 = 2x + 8 \\[3ex] x^2 - 2x - 8 = 0 \\[3ex] (x + 2)(x - 4) = 0 \\[3ex] x + 2 = 0 \;\;\;OR\;\;\; x - 4 = 0 \\[3ex] x = -2 \;\;\;OR\;\;\; x = 4 \\[3ex] $ Check
$x = -2 \;\;\;OR\;\;\; x = 4$
LHS RHS
$ x = -2 \\[3ex] 36 * 6^{2x + 6} \\[3ex] 36 * 6^{2(-2) + 6} \\[3ex] 36 * 6^{-4 + 6} \\[3ex] 36 * 6^{2} \\[3ex] 36 * 36 \\[3ex] 1296 \\[3ex] $
$ x = 4 \\[3ex] 36 * 6^{2x + 6} \\[3ex] 36 * 6^{2(4) + 6} \\[3ex] 36 * 6^{8 + 6} \\[3ex] 6^2 * 6^{14} \\[3ex] 6^{2 + 14} \\[3ex] 6^{16} $
$ x = -2 \\[3ex] 6^{x^2} \\[3ex] 6^{(-2)^2} \\[3ex] 6^4 \\[3ex] 1296 \\[3ex] $
$ x = 4 \\[3ex] 6^{x^2} \\[3ex] 6^{4^2} \\[3ex] 6^{16} $
(43.)


(44.) ACT What value of x makes the equation below true? $$ \dfrac{25^x}{5^2} = 5^4 $$ $ F.\;\; 3 \\[3ex] G.\;\; 6 \\[3ex] H.\;\; 8 \\[3ex] J.\;\; 25 \\[3ex] K.\;\; 625 \\[3ex] $

$ \dfrac{25^x}{5^2} = 5^4 \\[5ex] 25^x = 5^2 * 5^4 \\[3ex] (5^2)^x = 5^{2 + 4} \\[3ex] 5^{2x} = 5^6 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] 2x = 6 \\[3ex] x = \dfrac{6}{2} \\[5ex] x = 3 \\[3ex] $ Check
$x = 3$
LHS RHS
$ \dfrac{25^x}{5^2} \\[5ex] \dfrac{25^3}{5^2} \\[5ex] \dfrac{5^{2(3)}}{5^2} \\[5ex] \dfrac{5^6}{5^2} \\[5ex] 5^{6 - 2} \\[5ex] 5^4 $ $5^4$
(45.)


(46.) NZQA Solve the equation $8^{y} * 4^{y^2 - 8} = 16$


$ 8^{y}* 4^{y^2 - 8} = 16 \\[3ex] 2^{3(y)} * 2^{2(y^2 - 8)} = 2^4 \\[3ex] 2^{3y} * 2^{2y^2 - 16} = 2^4 \\[3ex] 2^{3y + (2y^2 - 16)} = 2^4 ...Law\;1...Exp \\[3ex] 2^{3y + 2y^2 - 16} = 2^4 \\[3ex] 2^{2y^2 + 3y - 16} = 2^4 \\[3ex] Same\;\;base;\;\;Equate\;\;the\;\;exponents \\[3ex] 2y^2 + 3y - 16 = 4 \\[3ex] 2y^2 + 3y - 16 - 4 = 0 \\[3ex] 2y^2 + 3y - 20 = 0 \\[3ex] 2y^2 + 8y - 5y - 20 = 0 \\[3ex] 2y(y + 4) - 5(y + 4) = 0 \\[3ex] y + 4 = 0 \;\;\;OR\;\;\; 2y - 5 = 0 \\[3ex] y = -4 \;\;\;OR\;\;\; 2y = 5 \\[3ex] y = -4 \;\;\;OR\;\;\; y = \dfrac{5}{2} \\[5ex] $ Check
$y = -4$ OR $x = \dfrac{5}{2}$
LHS RHS
$ y = -4 \\[3ex] 8^{y}* 4^{y^2 - 8} \\[3ex] 8^{-4} * 4^{(-4)^2 - 8} \\[3ex] 2^{3(-4)} * 4^{16 - 8} \\[3ex] 2^{-12} * 4^{8} \\[3ex] 2^{-12} * 2^{2(8)} \\[3ex] 2^{-12} * 2^{16} \\[3ex] 2^{-12 + 16}...Law\;1...Exp \\[3ex] 2^4 \\[3ex] 16 $ 16
$ y = \dfrac{5}{2} \\[5ex] 8^{y}* 4^{y^2 - 8} \\[3ex] 8^{\dfrac{5}{2}} * 4^{\left(\dfrac{5}{2}\right)^2 - 8} \\[5ex] 2^{3\left(\dfrac{5}{2}\right)} * 4^{\dfrac{25}{4} - \dfrac{32}{4}} \\[5ex] 2^{\dfrac{15}{2}} * 4^{-\dfrac{7}{4}} \\[5ex] 2^{\dfrac{15}{2}} * 2^{2\left(-\dfrac{7}{4}\right)} \\[5ex] 2^{\dfrac{15}{2} + -\dfrac{7}{2}}...Law\;1...Exp \\[5ex] 2^{\dfrac{15}{2} - \dfrac{7}{2}} \\[5ex] 2^{\dfrac{8}{2}} \\[5ex] 2^4 \\[3ex] 16 $ 16
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(48.)


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