Question

Question 1 (Multiple Choice)

Answer 1

Q1. The answer is 7 to the 5 over 4 power 
Let's write the fourth root of 7 to the fifth power as a radical. 7 to the fifth power is 7⁵. The fourth root of 7 to the fifth power is $\sqrt[4]{7^{5} }$.
Now, to rewrite it as a rational exponent, we will use the following:
$x^{ \frac{m}{n}}= \sqrt[n]{ x^{m} }$
Our radical is  $\sqrt[4]{7^{5} }$ which means that n = 4, m = 5.
So, the rational exponent it will be:
$\sqrt[4]{ 7^{5}} =7^{ \frac{5}{4}}$ which is the same as 7 to the 5 over 4 power


Q2. The answer is the eighth root of 2 to the fifth power.
Let's present 2 to the 7 over 8 power, all over 2 to the 1 over 4 power as a rational exponent.
2 to the 7 over 8 power is $2^{ \frac{7}{8}}$
2 to the 1 over 4 power is $2^{ \frac{1}{4} }$
2 to the 7 over 8 power, all over 2 to the 1 over 4 power is $\frac{2^{ \frac{7}{8}}}{2^{ \frac{1}{4} }}$
Using the rule: $\frac{x^{a} }{ x^{b} }= x^{a-b}$ we have:
$\frac{2^{ \frac{7}{8}}}{2^{ \frac{1}{4} }} =2^{ \frac{7}{8}- \frac{1}{4}} = 2^{ \frac{7}{8}- \frac{2}{8}}= 2^{ \frac{7-2}{8} } = 2^{ \frac{5}{8} }$
Since: $x^{ \frac{m}{n}}= \sqrt[n]{ x^{m} }$, then n = 8, m = 5
Therefore
$2^{ \frac{5}{8} = \sqrt[8]{ 2^{5} }$


Q3. The answer is 9 inches squared.
The area of the rectangle (A) is
A = l · w                (l - length, w - width).
It is given:
l = the cube root of 81 inches = $\sqrt[3]{81}= \sqrt[3]{3^{4} } =3^{ \frac{4}{3} }$
w = 3 to the 2 over 3 = $3^{ \frac{2}{3}}$

A = $3^{ \frac{4}{3} } * 3^{ \frac{2}{3} }$
Since: $x^{a} * x^{b}= x^{a+b}$ then:
A = $3^{ \frac{4}{3}+ \frac{2}{3}}= 3^{ \frac{4+2}{3} } = 3^{ \frac{6}{3} } = 3^{2}=9$


Q4. The answer is By simplifying 25 to 5² to make both powers base five and subtracting the exponents 
5 to the fourth power, over 25 = 52 is $\frac{ 5^{4}}{25}= 5^{2}$
Now, let's simplify 25 to 5²:
$\frac{ 5^{4}}{ 5^{2} }=5^{2}$
Since  $\frac{x^{a} }{ x^{b} }= x^{a-b}$, we will subtract the exponents:
$5^{4-2} = 5^{2}$
⇒ $5^{2} = 5^{2}$


Q5. The answer is the ninth root of 3 
3 to the 2 over 3 power is $3^{ \frac{2}{3} }$
3 to the 2 over 3 power, to the 1 over 6 power is $(3^{ \frac{2}{3} } } )^{ \frac{1}{6} }$
Since $(x^{a})^{b} =x ^{a*b}$ then:
$(3^{ \frac{2}{3}}) ^{ \frac{1}{6} } = 3^{ \frac{2}{3} * \frac{1}{6} } = 3^{ \frac{2}{18} }= 3^{ \frac{1}{9} }$
Since: $x^{ \frac{m}{n}}= \sqrt[n]{ x^{m} }$, then: n = 9, m = 1
$3^{ \frac{1}{9} }= \sqrt[9]{3^{1} } = \sqrt[9]{3}$