Question

9 which expression is equivalent to ( a^{\frac{3}{4}}b^{\frac{1}{2}} ), where ( a geq 0 ) and ( b geq 0 )? a) ( sqrt4{a^{3}b} ) b) ( sqrt4{a^{3}b^{2}} ) c) ( sqrt{a^{3}b} ) d) ( sqrt{a^{4}b^{2}} ) 10 which expression is equivalent to ( h^{\frac{5}{12}}(h^{-\frac{1}{4}})^{\frac{2}{3}} ), where ( h > 0 )? a) ( \frac{1}{h^{6}} ) b) ( sqrt{h^{5}} ) c) ( \frac{1}{sqrt6{h}} ) d) ( \frac{1}{sqrt5{h^{2}}} ) 11 ( sqrt8{41k}left( sqrt9{41k}
ight)^{12} ) for what value of ( x ) is the given expression equivalent to ( (41k)^{25x} ), where ( k > 1 )? 12 the expression ( 4^{18x} ) is equivalent to ( k^{3x} ), where ( k ) is a constant. what is the value of ( k )? 13 which expression is equivalent to ( 9^{\frac{1}{n}}(4^{\frac{1}{2n}}) ), where ( n ) is a positive integer? a) ( 24^{\frac{1}{n}} ) b) ( 12^{\frac{1}{n}} ) c) ( sqrtn{18} ) d) ( sqrtn{6} ) 14 ( \frac{10sqrt6{9^{3}x^{48}}}{sqrt4{6^{4}x}} ) the given expression is equivalent to ( ax^{b} ), where ( a > 0, b > 0 ), and ( x > 1 ). what is the value of ( ab )? 15 if ( m ) and ( n ) are both positive numbers, and ( 4m ) is equal to the cube root of the square of ( n ), for what value of ( x ) is ( m^{x} ) equal to ( n ) when ( m = 2 )? 16 if ( 2^{x + 3}-2^{x}=k(2^{x}) ), what is the value of ( k )? a) 3 b) 5 c) 7 d) 8

Explanation:

Problem 9

Step1: Convert exponents to roots

Use rule $x^{\frac{m}{n}}=\sqrt[n]{x^m}$. Rewrite each term:
$a^{\frac{3}{4}}=\sqrt[4]{a^3}$, $b^{\frac{1}{2}}=b^{\frac{2}{4}}=\sqrt[4]{b^2}$

Step2: Combine into single radical

Multiply terms under same root:
$\sqrt[4]{a^3} \cdot \sqrt[4]{b^2} = \sqrt[4]{a^3b^2}$

Step1: Simplify the exponent term

Multiply exponents: $(h^{-\frac{1}{4}})^{\frac{7}{3}}=h^{-\frac{1}{4} \cdot \frac{7}{3}}=h^{-\frac{7}{12}}$

Step2: Add exponents of like bases

$h^{\frac{5}{12}} \cdot h^{-\frac{7}{12}} = h^{\frac{5}{12}-\frac{7}{12}}=h^{-\frac{2}{12}}=h^{-\frac{1}{6}}$

Step3: Rewrite negative exponent

$h^{-\frac{1}{6}}=\frac{1}{h^{\frac{1}{6}}}=\frac{1}{\sqrt[6]{h}}$

Step1: Rewrite radicals as exponents

$\sqrt[8]{41k}=(41k)^{\frac{1}{8}}$, $(\sqrt[9]{41k})^{12}=(41k)^{\frac{12}{9}}=(41k)^{\frac{4}{3}}$

Step2: Add exponents of like bases

$(41k)^{\frac{1}{8}+\frac{4}{3}}=(41k)^{\frac{3+32}{24}}=(41k)^{\frac{35}{24}}$

Step3: Set equal to given form

$\frac{35}{24}=25x$ → $x=\frac{35}{24 \cdot 25}=\frac{7}{120}$

Answer:

B) $\boldsymbol{\sqrt[4]{a^3b^2}}$

---

Problem 10