Question

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Simplify 3 √7 divided by 5 √7

answer choices

7 to the power of 1 over 5

7 to the power of 8 over 15

7 to the power of 5 over 3

7 to the power of 2 over 15

Answer 1

Do you really mean

\[\large {{3\sqrt 7} \over {5\sqrt 7}} \]
?
Because none of the suggested answers are even close.

Answer 2

yes @LarsEighner

Answer 3

There is no power of 7 in the answer. Simply cancel the roots.

Answer 4

Here's the problem @LarsEighner

Answer 5

Oh. Okay. That is different.

\[ \large {{\sqrt[3] 7} \over {\sqrt[5] 7}}\]

can you express the radicals as exponents?

Answer 6

No, I don't even know where to begin.. all I know is that they both have the square root of 7 in them @larseighner

Answer 7

No. Neither of them has a square root. The numerator has a cube root and the denominator has a 5th root.

Now the square root of 7 is the same as 7 with an exponent of 1/2.

\[ \Large \sqrt 7 = 7^{1 \over 2} \]

So how can you represent the cube root (third root) of 7 as 7 with an exponent?

\[ \Large \sqrt[3] 7 = \]

Answer 8

I don't understand i'm confused?

Answer 9

We are trying to simplify by using the laws of exponents. You should not be getting this problem if you have never heard of the laws of exponents.

In exponents, roots are fractions.

\[\Large \sqrt 7 = 7^{1 \over 2}\]
\[\Large \sqrt[3] 7 = 7^{1 \over 3}\]
\[\Large \sqrt[4] 7 = 7^{1 \over 4}\]
\[\Large \sqrt [5]7 = 7^{1 \over 5}\]
\[\Large \sqrt [6]7 = 7^{1 \over 6}\]

So what eponent on 7 can replace \( \sqrt[3] 7 \) in the following:

\[\Large {{\sqrt[3] 7} \over {\sqrt[5] 7}} = \sqrt[3] 7 \cdot {1 \over {\sqrt[5] 7}} \]