Solved Examples on Exponential Equations

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For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
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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.) $x^7 = 128$

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?

$ A.\;\; 2 \\[3ex] B.\;\; 4 \\[3ex] C.\;\; 11 \\[3ex] D.\;\; 16 \\[3ex] E.\;\; 44 \\[3ex] $

$ 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

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(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 $	$16$

(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}} $

(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 $	$81$

(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

(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} $

(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}} $

(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 $	$255$

(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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(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} $

(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 $	$5^4$

(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

(50.) If $49^a = 7$ and $3^{a + b} = 81$, then b = ?

$ F.\;\; \dfrac{1}{2} \\[5ex] G.\;\; \dfrac{3}{2} \\[5ex] H.\;\; \dfrac{5}{2} \\[5ex] J.\;\; 3 \\[3ex] K.\;\; \dfrac{7}{2} \\[5ex] $

$ 49^a = 7 \\[3ex] (7^2)^a = 7^1 \\[3ex] 7^{2a} = 7^1 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] \implies \\[3ex] 2a = 1 \\[3ex] a = \dfrac{1}{2} \\[5ex] 3^{a + b} = 81 \\[3ex] 3^{a + b} = 3^4 \\[3ex] same\;\;base;\;\;equate\;\;exponents \\[3ex] \implies \\[3ex] a + b = 4 \\[3ex] b = 4 - a \\[3ex] b = 4 - \dfrac{1}{2} \\[5ex] b = \dfrac{8}{2} - \dfrac{1}{2} \\[5ex] b = \dfrac{8 - 1}{2} \\[5ex] b = \dfrac{7}{2} $

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