Question

Simplify. square root of 7 over fourth root of 7

7 to the power of negative 3 over 4
7 to the power of 1 over 4
7 to the power of 3 over 4
7

Answer 1

\[\sqrt[b]{x^a} = x^{\frac{ a }{ b }}\]
\[\frac{ x^a }{ x^b } = x^{a-b}\]

Answer 2

Use these rules.

Answer 3

i'm not sure how to get the problem to look like that, though.

Answer 4

Let's break it down then, what is the square root of 7 according to the first rule?

Answer 5

\(\Large {
\cfrac{\sqrt[2]{7}}{\sqrt[4]{7}}
\\ \quad \\ \quad \\
a^{\frac{{\color{blue} n}}{{\color{red} m}}} \implies \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad

\sqrt[{\color{red} m}]{a^{\color{blue} n}}\implies a^{\frac{{\color{blue} n}}{{\color{red} m}}}
\\\quad \\
% -----------------------------



\cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}}\qquad thus
\\ \quad \\
\cfrac{\sqrt[2]{7^1}}{\sqrt[4]{7^1}}\implies \cfrac{7^{\frac{1}{2}}}{7^{\frac{1}{4}}}\implies \cfrac{7^{\frac{1}{2}}}{1}\cdot \cfrac{1}{7^{\frac{1}{4}}}\implies 7^{\frac{1}{2}}\cdot 7^{-\frac{1}{4}}
}\)

Answer 6

Or just look at this very descriptive explanation.

Answer 7

so... simpify that... recall your exponent rules

Answer 8

it is a very good explanation. thank you.
how would i solve it ?

Answer 9

use the exponent rules, check your material on that
or just check this rules set http://ts3.mm.bing.net/th?id=HN.607986448326853914&pid=15.1&P=0

Answer 10

when multiplying two bases, different exponent
keep the base, sum the exponents

Answer 11

well
when multiplying two SAME bases, different exponent
keep the base, sum the exponents

Answer 12

so...i would be adding the exponents, then ? and i'd need to make the denominators the same ?

Answer 13

yes... since they're fractions
so you'd be adding the fractions, yes, so yo'd get the LCD
\(\large {
\cfrac{\sqrt[2]{7^1}}{\sqrt[4]{7^1}}\implies \cfrac{7^{\frac{1}{2}}}{7^{\frac{1}{4}}}\implies \cfrac{7^{\frac{1}{2}}}{1}\cdot \cfrac{1}{7^{\frac{1}{4}}}\implies 7^{\frac{1}{2}}\cdot 7^{-\frac{1}{4}}\implies 7^{\frac{1}{2}-\frac{1}{4}}
}\)

Answer 14

the lcd is 8, right ?

Answer 15

you could use 8, sure, is not the LCD, but doesn't matter since it'd be simplified anyway

Answer 16

oh, okay, so then the answer is 7^1/4, right ?

Answer 17

yeap