Question

@crispyrat

Answer 1

One of the ideas we want to use in this problem is how square roots are related to fractional exponents.

For example, \(\large\sqrt{x} = x^{\frac{1}{2}}\)

This also helps us understand a bit about how rationalizing denominators work.
\(\large x^{\frac{1}{2}}\times x^{\frac{1}{2}}=x^{\frac{1}{2}+\frac{1}{2}}=x^{1} = x\)

Answer 2

@snowflake0531 @az

Answer 3

This is why we are allowed to do something like:
\(\huge \frac{4}{\sqrt{5}}= \frac{4}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}} = \frac{4 \sqrt{5}}{5}\)

Answer 4

The situation becomes a little different with your ` 5th root ` in this problem
\(\large\sqrt[5]{x} = x^{\frac{1}{5}}\)

This means that the \(\sqrt[5]{16}\) in your problem is equal to \(16^{1/5}\)

In this case, the 16 is interesting, because it's prime factorization is a key point in this problem.

Answer 5

better use pemdas.......ask @Tranquility its the best way to do it

Answer 6

@Shawnte there's not a brain cell in your head that understands anything that I'm saying here, stay out of this

Answer 7

16 can be written as follows:
\(16 = 2\times2\times2\times2 = 2^4\)

This means that:
\(\large\sqrt[5]{16} = (16)^{\frac{1}{5}} = (2^4)^{\frac{1}{5}} = 2^{(4 \times \frac{1}{5})} = 2^{\frac{4}{5}}\)

Answer 8

Therefore, in order to "rationalize the denominator" of something with the 5th root of 16 in the bottom, we want to multiply the top and bottom of that fraction by \(2^{\frac{1}{5}}\)

Answer 9

I'll say that 4/5 = 0.8 and 1/5 = 0.2 for the sake of avoiding the weird fraction font in the exponents

\(\huge \frac{3}{\sqrt[5]{16}} = \frac{3}{2^{0.8}} = \frac{3}{2^{0.8}}\times\frac{2^{0.2}}{2^{0.2}} = \frac{3 \times 2^{0.2}}{2^{0.8}\times2^{0.2}}\)

\(= \huge \frac{3 \times 2^{(\frac{1}{5})}}{2^{(0.8+0.2)}} = \frac{3 ~\sqrt[5]{2}}{2^1} = \frac{3 ~\sqrt[5]{2}}{2}\)

Answer 10

You'll be able to notice what the ` a ` and ` b ` values for this problem should be at this point, but just let me know if you got confused at any part along the way.