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 [tex] \sqrt[4]{7^{5} } [/tex].
Now, to rewrite it as a rational exponent, we will use the following:
[tex] x^{ \frac{m}{n}}= \sqrt[n]{ x^{m} } [/tex]
Our radical is [tex] \sqrt[4]{7^{5} } [/tex] which means that n = 4, m = 5.
So, the rational exponent it will be:
[tex] \sqrt[4]{ 7^{5}} =7^{ \frac{5}{4}} [/tex] 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 [tex]2^{ \frac{7}{8}} [/tex]
2 to the 1 over 4 power is [tex] 2^{ \frac{1}{4} } [/tex]
2 to the 7 over 8 power, all over 2 to the 1 over 4 power is [tex] \frac{2^{ \frac{7}{8}}}{2^{ \frac{1}{4} }} [/tex]
Using the rule: [tex] \frac{x^{a} }{ x^{b} }= x^{a-b} [/tex] we have:
[tex]\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} } [/tex]
Since: [tex] x^{ \frac{m}{n}}= \sqrt[n]{ x^{m} } [/tex], then n = 8, m = 5
Therefore
[tex]2^{ \frac{5}{8} = \sqrt[8]{ 2^{5} } [/tex]
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 = [tex] \sqrt[3]{81}= \sqrt[3]{3^{4} } =3^{ \frac{4}{3} } [/tex]
w = 3 to the 2 over 3 = [tex]3^{ \frac{2}{3}} [/tex]
A = [tex] 3^{ \frac{4}{3} } * 3^{ \frac{2}{3} } [/tex]
Since: [tex]x^{a} * x^{b}= x^{a+b} [/tex] then:
A = [tex]3^{ \frac{4}{3}+ \frac{2}{3}}= 3^{ \frac{4+2}{3} } = 3^{ \frac{6}{3} } = 3^{2}=9 [/tex]
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 [tex] \frac{ 5^{4}}{25}= 5^{2} [/tex]
Now, let's simplify 25 to 5²:
[tex] \frac{ 5^{4}}{ 5^{2} }=5^{2} [/tex]
Since [tex] \frac{x^{a} }{ x^{b} }= x^{a-b} [/tex], we will subtract the exponents:
[tex]5^{4-2} = 5^{2} [/tex]
⇒ [tex] 5^{2} = 5^{2} [/tex]
Q5. The answer is the ninth root of 3
3 to the 2 over 3 power is [tex]3^{ \frac{2}{3} } [/tex]
3 to the 2 over 3 power, to the 1 over 6 power is [tex] (3^{ \frac{2}{3} } } )^{ \frac{1}{6} } [/tex]
Since [tex] (x^{a})^{b} =x ^{a*b} [/tex] then:
[tex] (3^{ \frac{2}{3}}) ^{ \frac{1}{6} } = 3^{ \frac{2}{3} * \frac{1}{6} } = 3^{ \frac{2}{18} }= 3^{ \frac{1}{9} } [/tex]
Since: [tex] x^{ \frac{m}{n}}= \sqrt[n]{ x^{m} } [/tex], then: n = 9, m = 1
[tex]3^{ \frac{1}{9} }= \sqrt[9]{3^{1} } = \sqrt[9]{3} [/tex]
