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# 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.

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.

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 $(a^3)^{21}$

$(a^3)^{21} \\[3ex] = a^{3 * 21} ...Law\:\: 5...Exp \\[3ex] = a^{63}$
(4.) ACT For all $a \ne 0$, $\dfrac{a^8}{a^4}$ is equivalent to

$A.\:\: 1 \\[3ex] B.\:\: a^2 \\[3ex] C.\:\: a^4 \\[3ex] D.\:\: a^{12} \\[3ex] E.\:\: a^{32} \\[3ex]$

$\dfrac{a^8}{a^4} \\[5ex] = a^{8 - 4} ...Law\;2...Exp \\[3ex] = a^4$
(5.) ACT If $\log_3{2} = p$ and $\log_3{5} = q$, which of the following expressions is equal to $10$?

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

$\log_3{2} = p \\[3ex] 2 = 3^p ...Relationship \\[3ex] \log_3{5} = q \\[3ex] 5 = 3^q ...Relationship \\[3ex] 2 * 5 = 10 \\[3ex] \implies 10 = 3^p * 3^q \\[3ex] 3^p * 3^q = 3^{p + q} ...Law\:\: 1...Exp$
(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 $xy =$ ?

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

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

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

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.) WASSCE: FM Simplify $\dfrac{625^{\left(\dfrac{3x}{4} - 1\right)} + 125^{(x - 1)}}{5^{(3x - 2)}}$

$\dfrac{625^{\left(\dfrac{3x}{4} - 1\right)} + 125^{(x - 1)}}{5^{(3x - 2)}} \\[7ex] = \dfrac{5^{4{\left(\dfrac{3x}{4} - 1\right)}} + 5^{3{(x - 1)}}}{5^{(3x - 2)}} \\[7ex] = \dfrac{5^{3x - 4} + 5^{3x - 3}}{5^{3x - 2}} \\[5ex] = \dfrac{5^{3x - 2}\left(\dfrac{5^{3x - 4}}{5^{3x - 2}} + \dfrac{5^{3x - 3}}{5^{3x - 2}}\right)}{5^{3x - 2}} \\[7ex] = \dfrac{5^{3x - 2}\left[5^{3x - 4 - (3x - 2)} + 5^{3x - 3 - (3x - 2)}\right]}{5^{3x - 2}} \\[5ex] = \dfrac{5^{3x - 2}\left[5^{3x - 4 - 3x + 2} + 5^{3x - 3 - 3x + 2}\right]}{5^{3x - 2}} \\[5ex] = \dfrac{5^{3x - 2}\left[5^{-2} + 5^{-1}\right]}{5^{3x - 2}} \\[5ex] = 5^{-2} + 5^{-1} \\[3ex] = \dfrac{1}{5^2} + \dfrac{1}{5^1} \\[5ex] = \dfrac{1}{25} + \dfrac{1}{5} \\[5ex] = \dfrac{1 + 5}{25} \\[5ex] = \dfrac{6}{25}$
(12.) CSEC Simplify $p^3q^2 * pq^5$

$p^3q^2 * pq^5 \\[3ex] = p^3 * q^2 * p * q^5 \\[3ex] = p^3 * p * q^2 * q^5 \\[3ex] = p^{3 + 1} * q^{2 + 5}...Law\;\;1...Exp \\[3ex] = p^4 *q^7 \\[3ex] = p^4q^7$
(13.)

$(a^3)^{21} \\[3ex] = a^{3 * 21} ...Law\:\: 5...Exp \\[3ex] = a^{63}$
(14.) 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

$\dfrac{(2a^{-1}\sqrt{b})^4}{ab^{-3}} \\[5ex] \sqrt{b} = b^{\dfrac{1}{2}} ...Law\:\: 7...Exp \\[3ex] = \dfrac{\left(2 * a^{-1} * b^{\dfrac{1}{2}}\right)^4}{a * b^{-3}} \\[5ex] = \dfrac{2^4 * (a^{-1})^4 * \left(b^{\dfrac{1}{2}}\right)^4}{a * b^{-3}} ...Law\:\: 5...Exp \\[5ex] 2^4 = 16 \\[3ex] (a^{-1})^4 = a^{-1 * 4} = a^{-4} ...Law\:\: 5...Exp \\[3ex] \left(b^{\dfrac{1}{2}}\right)^4 = b^{\dfrac{1}{2} * 4} = b^2 ...Law\:\: 5...Exp \\[3ex] = \dfrac{16 * a^{-4} * b^2}{a * b^{-3}} \\[5ex] = 16 * \dfrac{a^{-4}}{a^1} * \dfrac{b^2}{b^{-3}} \\[5ex] \dfrac{a^{-4}}{a^1} = a^{-4 - 1} = a^{-5} ...Law\:\: 2...Exp \\[5ex] a^{-5} = \dfrac{1}{a^5} ...Law\:\: 6...Exp \\[5ex] \dfrac{b^2}{b^{-3}} = b^{2 - (-3)} = b^{2 + 3} = b^5 ...Law\:\: 2...Exp \\[5ex] = 16 * \dfrac{1}{a^5} * b^5 \\[5ex] = \dfrac{16b^5}{a^5}$