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 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 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.) (a.) The diameter of the planet of Mars is approximately 6.8 × 10³ kilometers.
Express this number in standard notation.

(b.) Paul is an astronomer.
His infrared telescope is able to detect radiation with a wavelength of 0.00000384 meters.
Express this number in scientific notation.

$ (a.) \\[3ex] Diameter\;\;of\;\;Mars \\[3ex] \approx 6.8 * 10^3 \\[3ex] \approx 6.8 * 1000 \\[3ex] \approx 6800\;km ...standard\;\;notation \\[3ex] (b.) \\[3ex] Wavelength \\[3ex] = 0.00000384 \\[3ex] = 3.84 * 10^{-6}\;m ...scientific\;\;notation $

(2.) ACT Which of the following is equivalent to $(a^3)^{21}$

$ A.\;\; 63a \\[3ex] B.\;\; 24a \\[3ex] C.\;\; 3a^{21} \\[3ex] D.\;\; a^{24} \\[3ex] E.\;\; a^{63} \\[3ex] $

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

(3.) WASSCE Simplify: $\dfrac{3^{n - 1} * 27^{n + 1}}{81^n}$

$ A.\;\; 3^{2n} \\[3ex] B.\;\; 9 \\[3ex] C.\;\; 3^n \\[3ex] D.\;\; 3^{n + 1} \\[3ex] $

$ \dfrac{3^{n - 1} * 27^{n + 1}}{81^n} \\[5ex] \dfrac{3^{n - 1} * 3^{3(n + 1)}}{3^{4(n)}} \\[5ex] 3^{n - 1 + 3(n + 1) - 4(n)}...Laws\;1\;\;and\;\;2...Exp \\[3ex] 3^{n - 1 + 3n + 3 - 4n} \\[3ex] 3^{2} \\[3ex] 9 $

(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.) Simplify $3m \cdot 4m^3n^2 \cdot 5n^5$

DISSOCIATE – SOLVE – ASSOCIATE

$ 3m \cdot 4m^3n^2 \cdot 5n^5 \\[3ex] \underline{DISSOCIATE} \\[3ex] 3 \cdot m \cdot 4 \cdot m^3 \cdot n^2 \cdot 5 \cdot n^5 \\[3ex] \underline{SOLVE} \\[3ex] 3 \cdot 4 \cdot 5 \cdot m^1 \cdot m^3 \cdot n^2 \cdot n^5 \\[3ex] 60 \cdot m^{1 + 3} \cdot n^{2 + 5} ... Law\;1...Exp \\[3ex] \underline{ASSOCIATE} \\[3ex] 60 \cdot m^4 \cdot n^7 \\[3ex] 60m^4n^7 $

(6.) CSEC Simplify $p^3q^2 * pq^5$

DISSOCIATE – SOLVE – ASSOCIATE

$ p^3q^2 \cdot pq^5 \\[3ex] \underline{DISSOCIATE} \\[3ex] p^3 \cdot q^2 \cdot p \cdot q^5 \\[3ex] \underline{SOLVE} \\[3ex] p^3 \cdot p \cdot q^2 \cdot q^5 \\[3ex] p^{3 + 1} \cdot q^{2 + 5}...Law\;\;1...Exp \\[3ex] \underline{ASSOCIATE} \\[3ex] p^4 \cdot q^7 \\[3ex] p^4q^7 $

(7.) Simplify $\left(-5c^2d^4e^3\right)\left(-5c^3d^2e\right)^2$

DISSOCIATE – SOLVE – ASSOCIATE

$ \left(-5c^2d^4e^3\right)\left(-5c^3d^2e\right)^2 \\[3ex] \left(-5c^2d^4e^3\right) \cdot \left(-5c^3d^2e\right) \cdot \left(-5c^3d^2e\right) \\[3ex] \underline{DISSOCIATE} \\[3ex] -5 \cdot c^2 \cdot d^4 \cdot e^3 \cdot -5 \cdot c^3 \cdot d^2 \cdot e \cdot -5 \cdot c^3 \cdot d^2 \cdot e \\[3ex] \underline{SOLVE} \\[3ex] -5 \cdot -5 \cdot -5 \cdot c^2 \cdot c^3 \cdot c^3 \cdot d^4 \cdot d^2 \cdot d^2 \cdot e^3 \cdot e^1 \cdot e^1 \\[3ex] -125 \cdot c^{2 + 3 + 3} \cdot d^{4 + 2 + 2} \cdot e^{3 + 1 + 1} \\[3ex] \underline{ASSOCIATE} \\[3ex] -125 \cdot c^8 \cdot d^8 \cdot e^5 \\[3ex] -125c^8d^8e^5 $

(8.) Simplify $10xy^3 \cdot 8x^5y^3$

DISSOCIATE – SOLVE – ASSOCIATE

$ 10xy^3 \cdot 8x^5y^3 \\[3ex] \underline{DISSOCIATE} \\[3ex] 10 \cdot x \cdot y^3 \cdot 8 \cdot x^5 \cdot y^3 \\[3ex] \underline{SOLVE} \\[3ex] 10 \cdot 8 \cdot x \cdot x^5 \cdot y^3 \cdot y^3 \\[3ex] 80 \cdot x^{1 + 5} \cdot y^{3 + 3} ...Law\;1...Exp \\[3ex] 80 \cdot x^6 \cdot y^6 \\[3ex] \underline{ASSOCIATE} \\[3ex] 80x^6y^6 $

(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.) Simplify.
Write each answer in scientific notation

$ (a.)\;\; (6 * 10^6)(4 * 10^{-1}) \\[3ex] (b.)\;\; (4.11 * 10^5)(8.65 * 10^{-5}) \\[3ex] (c.)\;\; (8.31 * 10^{-3})(6.6 * 10^{-6}) \\[3ex] $

$ (a.) \\[3ex] (6 * 10^6)(4 * 10^{-1}) \\[3ex] 6 * 10^6 * 4 * 10^{-1} \\[3ex] 6 * 4 * 10^6 * 10^{-1} \\[3ex] 24 * 10^{6 + -1} ...Law\;1...Exp \\[3ex] 24 * 10^{5} \\[3ex] 2.4 * 10^{1} * 10^{5} \\[3ex] 2.4 * 10^{1 + 5} \\[3ex] 2.4 * 10^6 \\[5ex] (b.) \\[3ex] (4.11 * 10^5)(8.65 * 10^{-5}) \\[3ex] 4.11 * 10^5 * 8.65 * 10^{-5} \\[3ex] 4.11 * 8.65 * 10^5 * 10^{-5} \\[3ex] 35.5515 * 10^{5 + -5}...Law\;1...Exp \\[3ex] 35.5515 * 10^{0} \\[3ex] 35.5515 * 1 ...Law\;3...Exp \\[3ex] 35.5515 \\[3ex] 3.55515 * 10^1 \\[3ex] 3.55515 * 10 \\[5ex] (c.) \\[3ex] (8.31 * 10^{-3})(6.6 * 10^{-6}) \\[3ex] 8.31 * 10^{-3} * 6.6 * 10^{-6} \\[3ex] 8.31 * 6.6 * 10^{-3} * 10^{-6} \\[3ex] 54.846 * 10^{-3 + -6}...Law\;1...Exp \\[3ex] 5.4846 * 10^1 * 10^{-9} \\[3ex] 5.4846 * 10^{1 + -9} ...Law\;1...Exp \\[3ex] 5.4846 * 10^{-8} $

(11.) ACT Whenever $x$ and $y$ are nonzero,

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

DISSOCIATE – SOLVE – ASSOCIATE

$ \dfrac{(8x^5y^4)(6x^{13}y^3)}{16x^6y^{14}} \\[5ex] \underline{DISSOCIATE} \\[3ex] \dfrac{(8 \cdot x^5 \cdot y^4)(6 \cdot x^{13} \cdot y^3)}{16 \cdot x^6 \cdot y^{14}} \\[5ex] \underline{SOLVE} \\[3ex] \dfrac{8 \cdot 6}{16} \cdot \dfrac{x^5 \cdot x^{13}}{x^6} \cdot \dfrac{y^4 \cdot y^3}{y^{14}} \\[5ex] \dfrac{6}{2} \cdot x^{5 + 13 - 6} \cdot y^{4 + 3 - 14} ...Laws\: 1 \:\:and\:\: 2...Exp \\[5ex] 3 \cdot x^{12} * y^{-7} \\[3ex] 3 \cdot x^{12} * \dfrac{1}{y^7} ...Law\: 6...Exp \\[5ex] \underline{ASSOCIATE} \\[3ex] \dfrac{3x^{12}}{y^7} $

(12.) 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}} \\[5ex] $

$ (3 + x)^{-100} \\[3ex] = \dfrac{1}{(3 + x)^{100}} ...Law\: 6...Exp $

(13.) NYSED The expression $\left(x^{\dfrac{1}{2}}y^{-\dfrac{2}{3}}\right)^{-6}$ is equivalent to

$ (1)\;\; \dfrac{y^4}{x^3} \hspace{10em} (3)\;\; \dfrac{1}{x^3y^4} \\[5ex] (2)\;\; \dfrac{x^3}{y^4} \hspace{10em} (4)\;\; x^3y^4 \\[5ex] $

$ \left(x^{\dfrac{1}{2}}y^{-\dfrac{2}{3}}\right)^{-6} \\[5ex] x^{\dfrac{1}{2} * -6} \cdot y^{-\dfrac{2}{3} * -6} ...Law\;5...Exp \\[5ex] x^{-3} \cdot y^4 \\[3ex] \dfrac{1}{x^3} \cdot y^4 \\[5ex] \dfrac{y^4}{x^3} $

(14.) ACT The expression $(x^4)^6$ is equivalent to:

$ F.\;\; x^{10} \\[3ex] G.\;\; x^{24} \\[3ex] H.\;\; x^{4,096} \\[3ex] J.\;\; 6x^2 \\[3ex] K.\;\; 6x^3 \\[3ex] $

$ (x^4)^6 \\[3ex] x^{4 \cdot 6} ...Law\:\: 5...Exp \\[3ex] x^{24} $

(15.) NYSED The expression $\sqrt[3]{27a^{-6}b^3c^2}$ is equivalent to

$ (1)\;\; \dfrac{3bc^{\dfrac{2}{3}}}{a^2} \hspace{10em} (3)\;\; \dfrac{3b^6c^5}{a^3} \\[5ex] (2)\;\; \dfrac{3b^9c^6}{a^{18}} \hspace{10em} (4)\;\; \dfrac{3b\sqrt[3]{3c^2}}{a^2} \\[5ex] $

DISSOCIATE – SOLVE – ASSOCIATE

$ \sqrt[3]{27a^{-6}b^3c^2} \\[3ex] \underline{DISSOCIATE} \\[3ex] \sqrt[3]{27} \cdot \sqrt[3]{a^{-6}} \cdot \sqrt[3]{b^3} \cdot \sqrt[3]{c^2} \\[3ex] \underline{SOLVE} \\[3ex] 3 \cdot a^{-6 \left(\dfrac{1}{3}\right)} \cdot b^{3 \left(\dfrac{1}{3}\right)} \cdot c^{2 \left(\dfrac{1}{3}\right)}...Law\;7...Exp \\[5ex] 3 \cdot a^{-2} \cdot b^1 \cdot c^{\dfrac{2}{3}} \\[5ex] 3 \cdot \dfrac{1}{a^2} \cdot b \cdot c^{\dfrac{2}{3}}...Law\;6...Exp \\[5ex] \underline{ASSOCIATE} \\[3ex] \dfrac{3bc^{\dfrac{2}{3}}}{a^2} $

(16.) 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}} \\[3ex] $

$ \sqrt{\sqrt[3]{x^2}} \\[3ex] \sqrt[3]{x^2} = x^{\dfrac{2}{3}}...Law\: 7...Exp \\[3ex] = \sqrt{x^{\dfrac{2}{3}}} \\[5ex] = \left(x^{\dfrac{2}{3}}\right)^{\dfrac{1}{2}}...Law\;7...Exp \\[5ex] = x^{\dfrac{2}{3} * \dfrac{1}{2}}...Law\: 5...Exp \\[5ex] = x^{\dfrac{1}{3}} \\[5ex] $ 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 $

(17.) WASCCE Given that $t = 2^{-x}$, find $2^{x + 1}$ in terms of t

$ (A.)\;\; \dfrac{2}{t} \hspace{10em} (B.)\;\; \dfrac{t}{2} \\[5ex] (C.)\;\; \dfrac{1}{2t} \hspace{10em} (D.)\;\; 2t \\[5ex] $

$ t = 2^{-x} \\[3ex] 2^{x + 1} \\[3ex] = 2^x * 2^1 \\[3ex] = \dfrac{1}{2^{-x}} * 2 \\[5ex] = \dfrac{1}{t} * 2 \\[5ex] = \dfrac{2}{t} $

(18.) ACT Which of the following is equivalent to $(a^8)^{24}$

$ F.\;\; 192a \\[3ex] G.\;\; 32a \\[3ex] H.\;\; 8a^{24} \\[3ex] J.\;\; a^{32} \\[3ex] K.\;\; a^{192} \\[3ex] $

$ (a^8)^{24} \\[3ex] = a^{8 * 24} ...Law\:\: 5...Exp \\[3ex] = a^{192} $

(19.) WASSCE Simplify $\sqrt{\left(\dfrac{x^3y^5}{xy^7}\right)}$,
where $x \gt 0$, and $y \gt 0$

DISSOCIATE – SOLVE – ASSOCIATE

$ \sqrt{\left(\dfrac{x^3y^5}{xy^7}\right)} \\[5ex] = \sqrt{\left(\dfrac{x^3 \cdot y^5}{x \cdot y^7}\right)} \\[5ex] = \sqrt{\dfrac{x^3}{x} \cdot \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 \cdot y^{-2}} \\[3ex] = \sqrt{x^2} \cdot \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 \cdot \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 \cdot \dfrac{1}{2}} = y^{-1} ...Law\: 5...Exp \\[3ex] y^{-1} = \dfrac{1}{y} ...Law\: 6...Exp \\[5ex] = x \cdot \dfrac{1}{y} \\[5ex] = \dfrac{x}{y} $

(20.) NYSED Express $\dfrac{12x^{-5}y^5}{24x^{-3}y^{-2}}$ in simplest form, using only positive exponents.

DISSOCIATE – SOLVE – ASSOCIATE

$ \dfrac{12x^{-5}y^5}{24x^{-3}y^{-2}} \\[5ex] \dfrac{12}{24} \cdot \dfrac{x^{-5}}{x^{-3}} \cdot \dfrac{y^5}{y^{-2}} \\[5ex] \dfrac{1}{2} \cdot x^{-5 - (-3)} \cdot y^{5 - (-2)} \\[5ex] \dfrac{1}{2} \cdot x^{-5 + 3} \cdot y^{5 + 2} \\[5ex] \dfrac{1}{2} \cdot x^{-2} \cdot y^7 \\[5ex] \dfrac{1}{2} \cdot \dfrac{1}{x^2} \cdot y^7 \\[5ex] \dfrac{y^7}{2x^2} $

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(21.) NYSED The expression $\sqrt[4]{81x^2y^5}$ is equivalent to

$ (1)\;\; 3x^{\dfrac{1}{2}}y^{\dfrac{5}{4}} \hspace{10em} (3)\;\; 9xy^{\dfrac{5}{2}} \\[5ex] (2)\;\; 3x^{\dfrac{1}{2}}y^{\dfrac{4}{5}} \hspace{10em} (4)\;\; 9xy^{\dfrac{2}{5}} \\[5ex] $

DISSOCIATE – SOLVE – ASSOCIATE

$ \sqrt[4]{81x^2y^5} \\[3ex] \sqrt[4]{81} * \sqrt[4]{x^2} * \sqrt[4]{y^5} \\[3ex] 3 * x^{2\left(\dfrac{1}{4}\right)} * y^{5\left(\dfrac{1}{4}\right)} \\[5ex] 3 * x^{\dfrac{1}{2}} * y^{\dfrac{5}{4}} \\[5ex] 3x^{\dfrac{1}{2}}y^{\dfrac{5}{4}} $

(22.) CSEC Simplify the expression: $\dfrac{4x^2 * 3x^4}{6x^3}$

DISSOCIATE – SOLVE – ASSOCIATE

$ \dfrac{4x^2 \cdot 3x^4}{6x^3} \\[5ex] DISSOCIATE \\[3ex] \dfrac{4 \cdot x^2 \cdot 3 \cdot x^4}{6 \cdot x^3} \\[5ex] \dfrac{4 \cdot 3}{6} \cdot \dfrac{x^2 \cdot x^4}{x^3} \\[5ex] SOLVE \\[3ex] \dfrac{12}{6} \cdot \dfrac{x^{2 + 4}}{x^3} \\[5ex] 2 \cdot \dfrac{x^6}{x^3} \\[5ex] 2 \cdot x^{6 - 3} \\[3ex] 2 \cdot x^3 \\[3ex] ASSOCIATE \\[3ex] 2x^3 $

(23.) 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} $

(25.) WASCCE Simplify the expression $\dfrac{a^2b^4 - b^2a^4}{ab(a + b)}$

$ (A.)\;\; a^2 - b^2 \hspace{10em} (B.)\;\; b^2 - a^2 \\[3ex] (C.)\;\; a^2b - ab^2 \hspace{9em} (D.)\;\; ab^2 - a^2b \\[3ex] $

$ \dfrac{a^2b^4 - b^2a^4}{ab(a + b)} \\[5ex] \dfrac{a^2b^2(b^2 - a^2)}{ab(a + b)} \\[5ex] \dfrac{ab(b + a)(b - a)}{a + b} \\[5ex] \dfrac{ab(a + b)(b - a)}{a + b} \\[5ex] ab(b - a) \\[3ex] ab^2 - a^2b $

(26.) NYSED The expression $9^{\dfrac{3}{2}} \bullet 27^{\dfrac{1}{2}}$ is equivalent to

$ (1)\;\; 3^2 \hspace{10em} (3)\;\; 243^2 \\[5ex] (2)\;\; 3^{\dfrac{9}{2}} \hspace{10em} (4)\;\; 243^{\dfrac{3}{4}} \\[5ex] $

$ 9^{\dfrac{3}{2}} \bullet 27^{\dfrac{1}{2}} \\[5ex] 3^{2 * \dfrac{3}{2}} \bullet 3^{3 * \dfrac{1}{2}} \\[5ex] 3^3 \bullet 3^{\dfrac{3}{2}} \\[5ex] 3^{3 + \dfrac{3}{2}} \\[5ex] 3^{\dfrac{6}{2} + \dfrac{3}{2}} \\[5ex] 3^{\dfrac{9}{2}} $

(27.) NYSED The expression $\sqrt[4]{81x^8y^6}$ is equivalent to

$ (1)\;\; 3x^{2}y^{\dfrac{3}{2}} \hspace{10em} (3)\;\; 9x^2y^{\dfrac{3}{2}} \\[5ex] (2)\;\; 3x^{4}y^{2} \hspace{10em} (4)\;\; 9x^4y^{2} \\[5ex] $

DISSOCIATE – SOLVE – ASSOCIATE

$ \sqrt[4]{81x^8y^6} \\[3ex] \sqrt[4]{81} \cdot \sqrt[4]{x^8} \cdot \sqrt[4]{y^6} \\[3ex] 3 \cdot x^{8\left(\dfrac{1}{4}\right)} \cdot y^{6\left(\dfrac{1}{4}\right)} \\[5ex] 3 \cdot x^{2} \cdot y^{\dfrac{3}{2}} \\[5ex] 3x^{2}y^{\dfrac{3}{2}} $

(28.) Evaluate the expression: $\left(\dfrac{4}{5}\right)^{-3}$

Give the exact answer without an exponent.

$ \left(\dfrac{4}{5}\right)^{-3} \\[5ex] = 4^{-3} \div 5^{-3}...Law\;5...Exp \\[3ex] = \dfrac{1}{4^3} \div \dfrac{1}{5^3} ...Law\;6...Exp \\[5ex] = \dfrac{1}{64} \div \dfrac{1}{125} \\[5ex] = \dfrac{1}{64} * \dfrac{125}{1} \\[5ex] = \dfrac{125}{64} $

(29.) ACT Which of the following expressions is equivalent to $\left(x^5y^3z^2\right)\left(x^4y^3z^6\right)$ for all real values of x, y, and z?

$ F.\;\; x^9y^6z^8 \\[3ex] G.\;\; x^9y^9z^8 \\[3ex] H.\;\; x^{20}y^6z^8 \\[3ex] J.\;\; x^{20}y^9z^{12} \\[3ex] K.\;\; x^{21}y^6z^{12} \\[3ex] $

DISSOCIATE – SOLVE – ASSOCIATE

$ \left(x^5y^3z^2\right)\left(x^4y^3z^6\right) \\[3ex] x^5 \cdot y^3 \cdot z^2 \cdot x^4 \cdot y^3 \cdot z^6 \\[3ex] x^5 \cdot x^4 \cdot y^3 \cdot y^3 \cdot z^2 \cdot z^6 \\[3ex] x^{5 + 4} \cdot y^{3 + 3} \cdot z^{2 + 6} ...Law\;1...Exp \\[3ex] x^{9} \cdot y^6 \cdot z^8 \\[3ex] x^9y^6z^8 $

(30.) ACT For all nonzero values of x and y, which of the following expressions is equivalent to $-\dfrac{36x^4y^3}{4xy}$ ?

$ A.\;\; -40x^3y^2 \\[3ex] B.\;\; -32x^3y^2 \\[3ex] C.\;\; -9x^5y^4 \\[3ex] D.\;\; -9x^4y^3 \\[3ex] E.\;\; -9x^3y^2 \\[3ex] $

DISSOCIATE – SOLVE – ASSOCIATE

$ -\dfrac{36x^4y^3}{4xy} \\[5ex] = -1 \cdot \dfrac{36 \cdot x^4 \cdot y^3}{4 \cdot x \cdot y} \\[5ex] = -1 \cdot \dfrac{36}{4} \cdot \dfrac{x^4}{x^1} \cdot \dfrac{y^3}{y^1} \\[5ex] = -1 \cdot 9 \cdot x^{4 - 1} \cdot y^{3 - 1} ... Law\;2...Exp \\[3ex] = -9 \cdot x^3 \cdot y^2 \\[3ex] = -9x^3y^2 $

(31.) Simplify $\dfrac{\left(3x^5\right)^3}{\left(6x^2\right)^4}$ leaving your answer as positive exponents only.

We can solve this question using at least three approaches.
Use any approach you prefer.

$ \underline{First\;\;Approach:\;\;DISSOCIATE-SOLVE-ASSOCIATE} \\[3ex] \dfrac{\left(3x^5\right)^3}{\left(6x^2\right)^4} \\[5ex] = \dfrac{\left(3 \cdot x^5\right)^3}{\left(6 \cdot x^2\right)^4} \\[5ex] = \dfrac{3^3 \cdot x^{5 \cdot 3}}{6^4 \cdot x^{2 \cdot 4}} \\[5ex] = \dfrac{27 \cdot x^{15}}{1296 \cdot x^{8}} ...Law\;5...Exp \\[5ex] = \dfrac{27}{1296} \cdot \dfrac{x^{15}}{x^8} \\[5ex] = \dfrac{1}{48} \cdot x^{15 - 8} ...Law\;2...Exp \\[5ex] = \dfrac{1}{48} \cdot x^7 \\[5ex] = \dfrac{x^7}{48} \\[7ex] \underline{Second\;\;Approach:\;\;DISSOCIATE-SOLVE-ASSOCIATE} \\[3ex] \dfrac{\left(3x^5\right)^3}{\left(6x^2\right)^4} \\[5ex] = \dfrac{\left(3x^5\right)^3}{\left(6x^2\right)^3 \cdot (6x^2)^1...Law\;1...Exp} \\[5ex] = \left(\dfrac{3x^5}{6x^2}\right)^3 \cdot \dfrac{1}{6x^2} \\[5ex] = \left(\dfrac{3}{6} \cdot \dfrac{x^5}{x^2}\right)^3 \cdot \dfrac{1}{6x^2} \\[5ex] = \left(\dfrac{1}{2} \cdot x^{5 - 2}\right)^3 \cdot \dfrac{1}{6x^2} \\[5ex] = \left(\dfrac{1}{2} \cdot x^3\right)^3 \cdot \dfrac{1}{6x^2} \\[5ex] = \left(\dfrac{1}{2}\right)^3 \cdot (x^3)^3 \cdot \dfrac{1}{6x^2} \\[5ex] = \dfrac{1^3}{2^3} \cdot x^9 \cdot \dfrac{1}{6x^2} \\[5ex] = \dfrac{1}{8} \cdot \dfrac{x^9}{6x^2} \\[5ex] = \dfrac{1}{8 * 6} \cdot x^{9 - 2} \\[5ex] = \dfrac{1}{48} \cdot x^7 \\[5ex] = \dfrac{x^7}{48} \\[7ex] \underline{Third\;\;Approach:\;\;DISSOCIATE-SOLVE-ASSOCIATE} \\[3ex] \dfrac{\left(3x^5\right)^3}{\left(6x^2\right)^4} \\[5ex] = \dfrac{3x^5 \cdot 3x^5 \cdot 3x^5}{6x^2 \cdot 6x^2 \cdot 6x^2 \cdot 6x^2} \\[5ex] = \dfrac{3 \cdot x^5 \cdot 3 \cdot x^5 \cdot 3 \cdot x^5}{6 \cdot x^2 \cdot 6 \cdot x^2 \cdot 6 \cdot x^2 \cdot 6 \cdot x^2} \\[5ex] = \dfrac{3 \cdot 3 \cdot 3}{6 \cdot 6 \cdot 6 \cdot 6} \cdot \dfrac{x^{5 + 5 + 5}}{x^{2 + 2 + 2 + 2}} ...Law\;1...Exp \\[5ex] = \dfrac{1 \cdot 1 \cdot 1}{2 \cdot 2 \cdot 2 \cdot 6} \cdot \dfrac{x^{15}}{x^{8}} \\[5ex] = \dfrac{1}{48} \cdot x^{15 - 8} ... Law\;2...Exp \\[5ex] = \dfrac{1}{48} \cdot x^7 \\[5ex] = \dfrac{x^7}{48} $

(32.) Simplify $\dfrac{14x^4y^7}{6x^5y^4}$ leaving your answer as positive exponents only.

DISSOCIATE – SOLVE – ASSOCIATE

$ \dfrac{14x^4y^7}{6x^5y^4} \\[5ex] DISSOCIATE \\[3ex] = \dfrac{14}{6} \cdot \dfrac{x^4}{x^5} \cdot \dfrac{y^7}{y^4} \\[5ex] SOLVE \\[3ex] = \dfrac{7}{3} \cdot \dfrac{1}{x^{5 - 4}} \cdot \dfrac{y^{7 - 4}}{1} \\[5ex] = \dfrac{7}{3} \cdot \dfrac{1}{x} \cdot \dfrac{y^3}{1} \\[5ex] ASSOCIATE \\[3ex] = \dfrac{7y^3}{3x} $

(33.) JAMB Divide $a^{3x} - 26a^{2x} + 156a^x - 216$ by $a^{2x} - 24a^x + 108$

$ A.\:\: a^x - 2 \\[3ex] B.\:\: a^x + 2 \\[3ex] C.\:\: a^x - 18 \\[3ex] D.\:\: a^x - 6 \\[3ex] $

$ \begin{array}{c|c} & a^x - 2 \\ \hline a^{2x} - 24a^x + 108 & a^{3x} - 26a^{2x} + 156a^x - 216 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &a^{3x} - 24a^{2x} + 108a^x~~~~~~~~~~ \\ \hline &~~~~~~~-2a^{2x} + 48a^x - 216 \\ &-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \\ &~~~~~~~-2a^{2x} + 48a^x - 216 \\ \hline &~~~~~~~0 \\ \end{array} \\[5ex] Quotient = a^x - 2 $

(34.) ACT $4n^7 \cdot 3n^5$ is equivalent to

$ F.\;\; 7n^2 \\[3ex] G.\;\; 7n^{12} \\[3ex] H.\;\; 7n^{35} \\[3ex] J.\;\; 12n^{12} \\[3ex] K.\;\; 12n^{35} \\[3ex] $

DISSOCIATE – SOLVE – ASSOCIATE

$ 4n^7 \cdot 3n^5 \\[3ex] 4 * n^7 * 3 * n^5 \\[3ex] 4 * 3 * n^7 * n^5 \\[3ex] 12 * n^{7 + 5} ...Law\;1...Exp \\[3ex] 12 * n^{12} \\[3ex] 12n^{12} $

(35.) Simplify $5c^6 \cdot 3d^{-2} d \cdot 2e^{-8} c^{-5} e^{10}$ leaving your answer as positive exponents only.

DISSOCIATE – SOLVE – ASSOCIATE

$ 5c^6 \cdot 3d^{-2} d \cdot 2e^{-8} c^{-5} e^{10} DISSOCIATE \\[3ex] = 5 \cdot c^6 \cdot 3 \cdot d^{-2} \cdot d \cdot 2 \cdot e^{-8} \cdot c^{-5} \cdot e^{10} \\[3ex] SOLVE \\[3ex] = 5 \cdot 3 \cdot 2 \cdot c^6 \cdot c^{-5} \cdot d^{-2} \cdot d^1 \cdot e^{-8} \cdot e^{10} \\[3ex] = 30 \cdot c^{6 + -5} \cdot d^{-2 + 1} \cdot e^{-8 + 10} ...Law\;1...Exp \\[3ex] = 30 \cdot c^{6 - 5} \cdot d^{-1} \cdot e^{2} \\[3ex] = 30 \cdot c^1 \cdot \dfrac{1}{d} \cdot e^2 ...Law\;6...Exp \\[5ex] ASSOCIATE \\[3ex] = \dfrac{30 \cdot c \cdot e^2}{d} \\[5ex] = \dfrac{30ce^2}{d} $

(36.) Simplify $(2x^4y^4)^4$ leaving your answer as positive exponents only.

DISSOCIATE – SOLVE – ASSOCIATE

$ (2x^4y^4)^4 \\[3ex] DISSOCIATE \\[3ex] = (2 \cdot x^4 \cdot y^4)^4 \\[3ex] = 2^4 \cdot x^{4 * 4} \cdot y^{4 * 4} ...Law\;5...Exp \\[3ex] SOLVE \\[3ex] = 16 \cdot x^{16} \cdot y^{16} \\[3ex] ASSOCIATE \\[3ex] = 16x^{16}y^{16} $

(38.) 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] \implies \\[3ex] (4a)^{-1} \\[3ex] \dfrac{1}{4a} ...Law\:\:6...Exp $

(39.) ACT For all nonzero x, y, and z, which of the following is equal to $\left(\dfrac{2x^3y^{-5}z^8}{8x^{-2}y^6z^3}\right)^{-2} ?$

$ F.\;\; \dfrac{16x^{10}y^{22}}{z^{10}} \\[5ex] G.\;\; \dfrac{16y^{22}z^{10}}{x^{10}} \\[5ex] H.\;\; \dfrac{x^{10}y^{22}}{16z^{10}} \\[5ex] J.\;\; \dfrac{x^{10}z^{10}}{16y^{22}} \\[5ex] K.\;\; \dfrac{16y^{22}}{x^{10}z^{10}} \\[5ex] $

$ \left(\dfrac{2x^3y^{-5}z^8}{8x^{-2}y^6z^3}\right)^{-2} \\[5ex] Leave\;\;the\;\;outside\;\;exponent:\;\;-2 \\[3ex] We\;\;shall\;\;come\;\;to\;\;it\;\;later \\[3ex] \underline{Inside\;\;Parenthesis} \\[3ex] \dfrac{2x^3y^{-5}z^8}{8x^{-2}y^6z^3} \\[5ex] DISSOCIATE-SOLVE-ASSOCIATE \\[3ex] \implies \\[3ex] = \dfrac{2 \cdot x^3 \cdot y^{-5} \cdot z^8}{8 \cdot x^{-2} \cdot y^6 \cdot z^3} \\[5ex] = \dfrac{2}{8} \cdot \dfrac{x^3}{x^{-2}} \cdot \dfrac{y^{-5}}{y^6} \cdot \dfrac{z^8}{z^3} \\[5ex] = \dfrac{1}{4} \cdot x^{3 - (-2)} \cdot y^{-5 - 6} \cdot z^{8 - 3} ...Law\;2...Exp \\[5ex] = \dfrac{1}{4} \cdot x^5 \cdot y^{11} \cdot z^5 \\[5ex] Put\;\;back\;\;the\;\;outside\;\;exponent:\;\;-2 \\[3ex] \left(\dfrac{1}{4} \cdot x^5 \cdot y^{11} \cdot z^5\right)^{-2} \\[5ex] = \left(\dfrac{1}{4}\right)^2 \cdot x^{5(-2)} \cdot y^{-11(-2)} \cdot z^{5(-2)} ...Law\;5...Exp \\[5ex] = \dfrac{1}{\left(\dfrac{1}{4}\right)^2} \cdot x^{-10} \cdot y^{22} \cdot z^{-10} \\[7ex] = 1 \div \left(\dfrac{1}{4}\right)^2 \cdot \dfrac{1}{x^{10}} \cdot y^{22} \cdot \dfrac{1}{z^{10}} ...Law\;6...Exp \\[5ex] = 1 \div \dfrac{1^2}{4^2} \cdot \dfrac{y^{22}}{x^{10}z^{10}} \\[5ex] = 1 \div \dfrac{1}{16} \cdot \dfrac{y^{22}}{x^{10}z^{10}} \\[5ex] = 1 \cdot \dfrac{16}{1} \cdot \dfrac{y^{22}}{x^{10}z^{10}} \\[5ex] = \dfrac{16y^{22}}{x^{10}z^{10}} $

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(42.) 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 \\[3ex] = a^b \cdot c^b \\[3ex] = (ac)^b ...Law\:\: 5...Exp $

(43.) Simplify $\left(-4c^3d^2e^2\right)\left(4c^4de^3\right)^2$

DISSOCIATE – SOLVE – ASSOCIATE

$ \left(-4c^2d^2e^2\right)\left(4c^4de^3\right)^2 \\[3ex] \underline{DISSOCIATE} \\[3ex] -4 \cdot c^2 \cdot d^2 \cdot e^2 \cdot 4^2 \cdot c^{4(2)} \cdot d^{1(2)} \cdot e^{3(2)} ...Law\;5...Exp \\[3ex] \underline{SOLVE} \\[3ex] -4 \cdot c^2 \cdot d^2 \cdot e^2 \cdot 16 \cdot c^8 \cdot d^2 \cdot e^6 \\[3ex] -4 \cdot 16 \cdot c^2 \cdot c^8 \cdot d^2 \cdot d^2 \cdot e^2 \cdot e^6 \\[3ex] -64 \cdot c^{2 + 8} \cdot d^{2 + 2} \cdot e^{2 + 6} ...Law\;1..Exp \\[3ex] \underline{ASSOCIATE} \\[3ex] -64 \cdot c^{10} \cdot d^4 \cdot e^8 \\[3ex] -64c^{10}d^4e^8 $

(46.) ACT For all positive x and y, $x^{\dfrac{1}{3}}y^{\dfrac{5}{4}}$ can be written in which of the following radical forms?

$ F.\;\; \sqrt[12]{xy^5} \\[3ex] G.\;\; \sqrt[12]{x^4y^3} \\[3ex] H.\;\; \sqrt[12]{x^5y^5} \\[3ex] J.\;\; y\sqrt[12]{x^4y^3} \\[3ex] K.\;\; xy\sqrt[12]{x^7y^7} \\[3ex] $

$ x^{\dfrac{1}{3}}y^{\dfrac{5}{4}} \\[5ex] x^{\dfrac{1}{3}} * y^{\dfrac{5}{4}} \\[5ex] x^{\dfrac{1}{3}} * y^{\dfrac{4}{4}} * y^{\dfrac{1}{4}} ...Law\;1...Exp \\[5ex] x^{\dfrac{1}{3}} * y^1 * y^{\dfrac{1}{4}} \\[5ex] x^{\dfrac{1}{3}} * y * y^{\dfrac{1}{4}} \\[5ex] \sqrt[3]{x} * y * \sqrt[4]{y}...Law\;7...Exp \\[3ex] y * \sqrt[3]{x} * \sqrt[4]{y} \\[3ex] $ This is not the option.
Looking at the options, it is important to express the radical as the 12-th root rather than the 3rd-root for the x and the 4th-root for the y

$ \sqrt[12]{something} = something^{\dfrac{1}{12}}...Law\;7...Exp \\[5ex] For\;\;the\;\;x\;\;(3rd-root\;\;to\;\;12-root) \\[3ex] \dfrac{1}{12} * what = \dfrac{1}{3} \\[5ex] what = \dfrac{1}{3} * 12 \\[5ex] what = 4 \\[3ex] x^{what} = x^4 \\[3ex] For\;\;the\;\;y\;\;(4th-root\;\;to\;\;12-root) \\[3ex] \dfrac{1}{12} * what = \dfrac{1}{4} \\[5ex] what = \dfrac{1}{4} * 12 \\[5ex] what = 3 \\[3ex] y^{what} = y^3 \\[3ex] \implies \\[3ex] y * \sqrt[3]{x} * \sqrt[4]{y} \\[3ex] y * \sqrt[12]{x^4} * \sqrt[12]{y^3} \\[3ex] y * \sqrt[12]{x^4 * y^3} \\[3ex] y\sqrt[12]{x^4y^3} \\[5ex] \underline{Check} \\[3ex] y\sqrt[12]{x^4y^3} \\[3ex] y * \sqrt[12]{x^4 * y^3} \\[3ex] y * \sqrt[12]{x^4} * \sqrt[12]{y^3} \\[3ex] y * x^{4\left(\dfrac{1}{12}\right)} * y^{3\left(\dfrac{1}{12}\right)} \\[5ex] y * x^{\dfrac{1}{3}} * y^{\dfrac{1}{4}} \\[5ex] x^{\dfrac{1}{3}} * y^1 * y^{\dfrac{1}{4}} \\[5ex] x^{\dfrac{1}{3}} * y^{1 + \dfrac{1}{4}} \\[5ex] x^{\dfrac{1}{3}} * y^{\dfrac{4}{4} + \dfrac{1}{4}} \\[5ex] x^{\dfrac{1}{3}} * y^{\dfrac{5}{4}} \\[5ex] x^{\dfrac{1}{3}}y^{\dfrac{5}{4}} $

(47.) Simplify the expression. Use only positive exponents in your answer.

$6c^6c^{-3}\cdot 2a^{-8}d^{-10}\cdot 3ad^9$

DISSOCIATE – SOLVE – ASSOCIATE

$ 6c^6c^{-3}\cdot 2a^{-8}d^{-10}\cdot 3ad^9 \\[3ex] \underline{DISSOCIATE} \\[3ex] 6 \cdot c^6 \cdot c^{-3} \cdot 2 \cdot a^{-8} \cdot d^{-10} \cdot 3 \cdot a \cdot d^9 \\[3ex] \underline{SOLVE} \\[3ex] 6 \cdot 2 \cdot 3 \cdot c^6 \cdot c^{-3} \cdot a^{-8} \cdot a^1 \cdot d^{-10} \cdot d^9 \\[3ex] 36 \cdot c^{6 + (-3)} \cdot a^{-8 + 1} \cdot d^{-10 + 9} \\[3ex] 36 \cdot c^3 \cdot a^{-7} \cdot d^{-1} \\[3ex] 36 \cdot c^3 \cdot \dfrac{1}{a^7} \cdot \dfrac{1}{d} \\[5ex] \underline{ASSOCIATE} \\[3ex] \dfrac{36c^3}{a^7d} $

(50.) ACT Which of the following expressions is equivalent to $\sqrt[4]{256x^{16}}$

$ F.\;\; 4x^4 \\[3ex] G.\;\; 4x^{12} \\[3ex] H.\;\; 16x^4 \\[3ex] J.\;\; 64x^{12} \\[3ex] K.\;\; 128x^8 \\[3ex] $

$ \sqrt[4]{256x^{16}} \\[3ex] = \sqrt[4]{256 \cdot x^{16}} \\[3ex] = \sqrt[4]{256} \cdot \sqrt[4]{x^{16}} \\[3ex] = 4 \cdot x^{16 * \dfrac{1}{4}}...Law\;7...Exp \\[5ex] = 4 \cdot x^4 \\[3ex] = 4x^4 $

(52.) ACT For all nonzero values of x and y, which of the following expressions is equivalent to $-\dfrac{28x^4y^3}{4xy}$ ?

$ F.\;\; -7x^3y^2 \\[3ex] G.\;\; -7x^4y^4 \\[3ex] H.\;\; -7x^5y^4 \\[3ex] J.\;\; -24x^3y^2 \\[3ex] K.\;\; -32x^3y^2 \\[3ex] $

DISSOCIATE – SOLVE – ASSOCIATE

$ -\dfrac{28x^4y^3}{4xy} \\[5ex] = -1 \cdot \dfrac{28 \cdot x^4 \cdot y^3}{4 \cdot x \cdot y} \\[5ex] = -1 \cdot \dfrac{28}{4} \cdot \dfrac{x^4}{x^1} \cdot \dfrac{y^3}{y^1} \\[5ex] = -1 \cdot 7 \cdot x^{4 - 1} \cdot y^{3 - 1} ... Law\;2...Exp \\[3ex] = -7 \cdot x^3 \cdot y^2 \\[3ex] = -7x^3y^2 $

(54.) ACT If a and b are positive real numbers, which of the following is equivalent to $\dfrac{(2a^{-1}\sqrt{b})^4}{ab^{-3}}$

$ A.\;\; 8a^2b^4 \\[3ex] B.\;\; \dfrac{8b^6}{a^4} \\[5ex] C.\;\; \dfrac{16b^5}{a^5} \\[5ex] D.\;\; \dfrac{16b^4}{a^5} \\[5ex] E.\;\; \dfrac{16b}{a^3} \\[5ex] $

DISSOCIATE – SOLVE – ASSOCIATE

$ \dfrac{(2a^{-1}\sqrt{b})^4}{ab^{-3}} \\[5ex] \dfrac{\left(2a^{-1}b^{\dfrac{1}{2}}\right)^4 ...Law\;7...Exp}{ab^{-3}} \\[5ex] \underline{DISSOCIATE} \\[3ex] \dfrac{2^4 \cdot a^{-1(4)} \cdot b^{\dfrac{1}{2}(4)}}{a \cdot b^{-3}} \\[5ex] \underline{SOLVE} \\[3ex] \dfrac{16 \cdot a^{-4} \cdot b^2}{a \cdot b^{-3}} \\[5ex] 16 \cdot \dfrac{a^{-4}}{a^1} \cdot \dfrac{b^2}{b^{-3}} \\[5ex] 16 \cdot a^{-4 - 1} \cdot b^{2 - (-3)}...Law\;2...Exp \\[3ex] 16 \cdot a^{-5} \cdot b^5 \\[3ex] 16 \cdot \dfrac{1}{a^5} \cdot b^5 ...Law\;6...Exp \\[3ex] \underline{ASSOCIATE} \\[3ex] \dfrac{16b^5}{a^5} $

(56.) ACT Which of the expressions represents the sum of $3.8 \times 10^5$ and $6.4 \times 10^4$ in scientific notation?

$ F.\;\; 1.02 \times 10^{10} \\[3ex] G.\;\; 4.44 \times 10^4 \\[3ex] H.\;\; 4.44 \times 10^5 \\[3ex] J.\;\; 10.2 \times 10^{20} \\[3ex] K.\;\; 44.4 \times 10^4 \\[3ex] $

$ (3.8 \times 10^5) + (6.4 \times 10^4) \\[3ex] 3.8 \times 10^5 + 6.4 \times 10^4 \\[3ex] GCF = 10^4 \\[3ex] \implies \\[3ex] 10^4(3.8 \times 10 + 6.4) \\[3ex] 10^4(38 + 6.4) \\[3ex] 10^4 \times 44.4 \\[3ex] 44.4 \times 10^4 \\[3ex] Expressing\;\;in\;\;scientific\;\;notation \\[3ex] 4.44 \times 10 \times 10^4 \\[3ex] 4.44 \times 10^{1 + 4} ...Law\;1...Exp \\[3ex] 4.44 \times 10^5 $

(60.) ACT For all positive values of x, which of the following expressions is equivalent to $\sqrt[6]{x^4}(\sqrt[3]{x^4})$?

$ F.\;\; x^{\dfrac{8}{9}} \\[5ex] G.\;\; x \\[3ex] H.\;\; x^{\dfrac{9}{8}} \\[5ex] J.\;\; x^2 \\[3ex] K.\;\; x^{\dfrac{9}{4}} \\[5ex] $

$ \sqrt[6]{x^4}(\sqrt[3]{x^4}) \\[3ex] \sqrt[6]{x^4} \cdot (\sqrt[3]{x^4}) \\[3ex] (x^4)^{\dfrac{1}{6}} \cdot (x^4)^{\dfrac{1}{3}} ...Law\;7...Exp \\[5ex] x^{\dfrac{4}{6}} \cdot x^{\dfrac{4}{3}}...Law\;5...Exp \\[5ex] x^{\dfrac{2}{3}} \cdot x^{\dfrac{4}{3}} \\[5ex] x^{\dfrac{2}{3} + \dfrac{4}{3}}...Law\;1...Exp \\[5ex] x^{\dfrac{2 + 4}{3}} \\[5ex] x^{\dfrac{6}{3}} \\[5ex] x^2 $

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(68.) ACT Which of the following expressions is equivalent to $x^{\dfrac{2}{3}}$

$ F.\;\; \dfrac{x^2}{3} \\[5ex] G.\;\; \dfrac{x(2)}{3} \\[5ex] H.\;\; \sqrt{x^3} \\[3ex] J.\;\; \sqrt[3]{x} \\[3ex] K.\;\; \sqrt[3]{x^2} \\[3ex] $

$ x^{\dfrac{2}{3}} \\[5ex] = \sqrt[3]{x^2}...Law\;7...Exp $