Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka

Solved Examples on Exponential Expressions

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

Solve each exponential expression.
Indicate the law(s) used.
Show all Work.

(1.) ACT Which of the following expressions is equivalent to $(3 + x)^{-100}$

$ A.\: -3^{100} - x{100} \\[3ex] B.\: -300 - 100x \\[3ex] C.\: \dfrac{1}{3^{300}} + \dfrac{1}{x^{100}} \\[5ex] D.\: \dfrac{1}{(3x)^{100}} \\[5ex] E.\: \dfrac{1}{(3 + x)^{100}} $


$ (3 + x)^{-100} \\[3ex] = \dfrac{1}{(3 + x)^{100}} ...Law\: 6...Exp $
(2.) ACT For all $x \gt 0$, which of the following expressions is NOT equivalent to: $\sqrt{\sqrt[3]{x^2}}$

$ F.\:\: \sqrt[3]{x} \\[3ex] G.\:\: \sqrt[6]{x^2} \\[3ex] H.\:\: \sqrt[3]{\sqrt{x^2}} \\[3ex] J.\:\: x^{\dfrac{1}{3}} \\[3ex] K.\:\: x^{\dfrac{2}{3}} $


$ \sqrt{\sqrt[3]{x^2}} \\[3ex] \sqrt[3]{x^2} = x^{\dfrac{2}{3}}...Law\: 7...Exp \\[3ex] = \sqrt{x^{\dfrac{2}{3}}} \\[3ex] = (x^{\dfrac{2}{3}})^{\dfrac{1}{2}} \\[3ex] = x^{\dfrac{2}{3} * \dfrac{1}{2}}...Law\: 5...Exp \\[3ex] = x^{\dfrac{1}{3}} \\[3ex] $ Let us solve each option and see which one is not equivalent to $x^{\dfrac{1}{3}}$

$ F.\:\: \sqrt[3]{x} \\[3ex] \sqrt[3]{x} = x^{\dfrac{1}{3}}...Law\: 7...Exp \\[3ex] YES \\[5ex] G.\:\: \sqrt[6]{x^2} \\[3ex] \sqrt[6]{x^2} = (x^2)^{\dfrac{1}{6}}...Law\: 7...Exp \\[3ex] = x^{2 * \dfrac{1}{6}}...Law\: 5...Exp \\[3ex] = x^{\dfrac{1}{3}} \\[3ex] YES \\[5ex] H.\:\: \sqrt[3]{\sqrt{x^2}} \\[3ex] \sqrt{x^2} = (x^2)^{\dfrac{1}{2}}...Law\: 7...Exp \\[3ex] \sqrt{x^2} = x^{2 * \dfrac{1}{2}}...Law\: 5...Exp \\[3ex] \sqrt{x^2} = x \\[3ex] = \sqrt[3]{x} = x^{\dfrac{1}{3}}...Law\: 7...Exp \\[3ex] YES \\[5ex] J.\:\: x^{\dfrac{1}{3}} \\[3ex] YES \\[5ex] K.\:\: x^{\dfrac{2}{3}} \\[3ex] NO $
(3.) ACT Which of the following is equivalent to $(a^3)^{21}$


$ (a^3)^{21} = a^{3 * 21} ...Law\:\: 5...Exp \\[3ex] = a^{63} $
(4.) ACT If $a$ and $b$ are positive real numbers, which of the following is equivalent to $\dfrac{(2a^{-1}\sqrt{b})^4}{ab^{-3}}$


We shall use the method: DISSOCIATE - SOLVE - ASSOCIATE

$ \dfrac{(2a^{-1}\sqrt{b})^4}{ab^{-3}} \\[5ex] \sqrt{b} = b^{\dfrac{1}{2}} ...Law\:\: 7...Exp \\[3ex] = \dfrac{\left(2 * a^{-1} * b^{\dfrac{1}{2}}\right)^4}{a * b^{-3}} \\[5ex] = \dfrac{2^4 * (a^{-1})^4 * \left(b^{\dfrac{1}{2}}\right)^4}{a * b^{-3}} ...Law\:\: 5...Exp \\[5ex] 2^4 = 16 \\[3ex] (a^{-1})^4 = a^{-1 * 4} = a^{-4} ...Law\:\: 5...Exp \\[3ex] \left(b^{\dfrac{1}{2}}\right)^4 = b^{\dfrac{1}{2} * 4} = b^2 ...Law\:\: 5...Exp \\[3ex] = \dfrac{16 * a^{-4} * b^2}{a * b^{-3}} \\[5ex] = 16 * \dfrac{a^{-4}}{a^1} * \dfrac{b^2}{b^{-3}} \\[5ex] \dfrac{a^{-4}}{a^1} = a^{-4 - 1} = a^{-5} ...Law\:\: 2...Exp \\[5ex] a^{-5} = \dfrac{1}{a^5} ...Law\:\: 6...Exp \\[5ex] \dfrac{b^2}{b^{-3}} = b^{2 - (-3)} = b^{2 + 3} = b^5 ...Law\:\: 2...Exp \\[5ex] = 16 * \dfrac{1}{a^5} * b^5 \\[5ex] = \dfrac{16b^5}{a^5} $
(5.) ACT If $\log_3{2} = p$ and $\log_3{5} = q$, which of the following expressions is equal to $10$?

$ F.\:\: 3^{p + q} \\[3ex] G.\:\: 3^p + 3^q \\[3ex] H.\:\: 9^{p + q} \\[3ex] J.\:\: pq \\[3ex] K.\:\: p + q $


$ \log_3{2} = p \\[3ex] 2 = 3^p ...Relationship \\[3ex] \log_3{5} = q \\[3ex] 5 = 3^q ...Relationship \\[3ex] 2 * 5 = 10 \\[3ex] \implies 10 = 3^p * 3^q \\[3ex] 3^p * 3^q = 3^{p + q} ...Law\:\: 1...Exp $
(6.) ACT Whenever $x$ and $y$ are nonzero,

$\dfrac{(8x^5y^4)(6x^{13}y^3)}{16x^6y^{14}} = ?$


$ \dfrac{(8x^5y^4)(6x^{13}y^3)}{16x^6y^{14}} \\[5ex] DISSOCIATE-SOLVE-ASSOCIATE \\[3ex] = \dfrac{(8 * x^5 * y^4)(6 * x^{13} * y^3)}{16 * x^6 * y^{14}} \\[5ex] = \dfrac{8 * 6}{16} * \dfrac{x^5 * x^{13}}{x^6} * \dfrac{y^4 * y^3}{y^{14}} \\[5ex] \dfrac{8 * 6}{16} = \dfrac{6}{2} = 3 \\[5ex] \dfrac{x^5 * x^{13}}{x^6} = x^{5 + 13 - 6} = x^{12} ...Laws\:\: 1 \:\:and\:\: 2...Exp \\[5ex] \dfrac{y^4 * y^3}{y^{14}} = y^{4 + 3 - 14} = y^{-7} ...Laws\:\: 1 \:\:and\:\: 2...Exp \\[5ex] y^{-7} = \dfrac{1}{y^7} ...Law\:\: 6...Exp \\[5ex] = 3 * x^{12} * \dfrac{1}{y^7} \\[5ex] = \dfrac{3x^{12}}{y^7} $
(7.) WASSCE Simplify $\sqrt{\left(\dfrac{x^3y^5}{xy^7}\right)}$,
where $x \gt 0$, and $y \gt 0$


$ \sqrt{\left(\dfrac{x^3y^5}{xy^7}\right)} \\[5ex] DISSOCIATE-SOLVE-ASSOCIATE \\[3ex] = \sqrt{\left(\dfrac{x^3 * y^5}{x * y^7}\right)} \\[5ex] = \sqrt{\dfrac{x^3}{x} * \dfrac{y^5}{y^7}} \\[5ex] \dfrac{x^3}{x} = x^{3 - 1} = x^2 ...Law\: 2...Exp \\[5ex] \dfrac{y^5}{y^7} = y^{5 - 7} = y^{-2} ...Law\: 2...Exp \\[5ex] = \sqrt{x^2 * y^{-2}} \\[3ex] = \sqrt{x^2} * \sqrt{y^{-2}} \\[3ex] \sqrt{x^2} = \left(x^2\right)^\dfrac{1}{2} ...Law\: 7...Exp \\[3ex] \left(x^2\right)^\dfrac{1}{2} = x^{2 * \dfrac{1}{2}} = x ...Law\: 5...Exp \\[3ex] \sqrt{y^{-2}} = \left(y^{-2}\right)^\dfrac{1}{2} ...Law\: 7...Exp \\[3ex] \left(y^{-2}\right)^\dfrac{1}{2} = y^{-2 * \dfrac{1}{2}} = y^{-1} ...Law\: 5...Exp \\[3ex] y^{-1} = \dfrac{1}{y} ...Law\: 6...Exp \\[5ex] = x * \dfrac{1}{y} \\[5ex] = \dfrac{x}{y} $
(8.) ACT If $a$, $b$, and $c$ are positive integers such that $a^b = x$ and $c^b = y$, then $xy =$ ?

$ A.\:\: ac^b \\[3ex] B.\:\: ac^{2b} \\[3ex] C.\:\: (ac)^b \\[3ex] D.\:\: (ac)^{2b} \\[3ex] E.\:\: (ac)^{b^2} $


$ a^b = x \\[3ex] c^b = y \\[3ex] xy = a^b * c^b = (ac)^b ...Law\:\: 5...Exp $
(9.) ACT For positive real numbers $x$, $y$, and $z$, which of the following expressions is equivalent to $x^{\dfrac{1}{2}}y^{\dfrac{2}{3}}z^{\dfrac{5}{6}}$?

$ A.\:\: \sqrt[3]{xy^2z^3} \\[3ex] B.\:\: \sqrt[6]{xy^2z^5} \\[3ex] C.\:\: \sqrt[6]{x^3y^2z^5} \\[3ex] D.\:\: \sqrt[6]{x^3y^4z^5} \\[3ex] E.\:\: \sqrt[11]{xy^2z^5} $


The ACT is a timed test where each question is expected to be done in a minute "at most".
Some of the questions can be done in less than a minute.
Some of the questions may take more than a minute.
The expectation is that you should use the "extra" time saved from those questions that were done in less than a minute on the questions that would take more than a minute.
Time management skills is being tested.
Let us look at the denominators - $2, 3, 6$
The $LCD = 6$
So, we should be looking at the $6th$ root options.
the $6th$ root is the fraction - $\dfrac{1}{6}...Law\:\: 7...Exp$

How do we express the three exponents $\dfrac{1}{2}, \dfrac{2}{3}, \dfrac{5}{6}$ in terms of $\dfrac{1}{6}$

$ \dfrac{1}{6} * 3 = \dfrac{1}{2}...x \\[5ex] \dfrac{1}{6} * 4 = \dfrac{2}{3}...y \\[5ex] \dfrac{1}{6} * 5 = \dfrac{1}{2}...z \\[5ex] \therefore x^{\dfrac{1}{2}}y^{\dfrac{2}{3}}z^{\dfrac{5}{6}} = \sqrt[6]{x^3y^4z^5} \\[3ex] \underline{Check} \\[3ex] \sqrt[6]{x^3} = x^{3 * \dfrac{1}{6}} = x^{\dfrac{1}{2}} \\[5ex] \sqrt[6]{y^4} = y^{4 * \dfrac{1}{6}} = y^{\dfrac{2}{3}} \\[5ex] \sqrt[6]{z^5} = z^{5 * \dfrac{1}{6}} = z^{\dfrac{5}{6}} \\[5ex] $
(10.) JAMB Simplify $\left(\sqrt[3]{64a^3}\right)^{-1}$

$ A.\:\: 4a \\[3ex] B.\:\: \dfrac{1}{8a} \\[5ex] C.\:\: 8a \\[3ex] D.\:\: \dfrac{1}{4a} \\[5ex] $

$ \left(\sqrt[3]{64a^3}\right)^{-1} \\[3ex] \sqrt[3]{64a^3} = \sqrt[3]{64} * \sqrt[3]{a^3} \\[3ex] \sqrt[3]{64a^3} = 4a \\[3ex] \rightarrow (4a)^{-1} \\[3ex] (4a)^{-1} = \dfrac{1}{4a} ...Law\:\:6...Exp $
(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 $