Remember that fraction exponents are the same as radicals.

\(\displaystyle \small 16^{\frac{1}{2}}=\sqrt{16}=4\)

\(\displaystyle 256^{\frac{3}{4}}=\sqrt[4]{256^3}=64\)

A shortcut would be to express the terms as exponents and look for opportunities to cancel.

\(\displaystyle 16^{\frac{1}{2}}=(4^2)^{\frac{1}{2}}=4\)

\(\displaystyle \small 256^{\frac{3}{4}}=(4^4)^{\frac{3}{4}}=4^3=64\)

Either method, we then need to multiply to two terms.

\(\displaystyle \small (4)(64)=256\)

Remember that exponents in the denominator refer to the root of the term, while exponents in the numerator can be treated normally.

\(\displaystyle x^{\frac{a}{b}}=\sqrt[b]{x^a}\)

\(\displaystyle x^{\frac{3}{7}}=\sqrt[7]{x^3}\)

\(\displaystyle [x^{m}]^{n}=x^{m\times n}\)

\(\displaystyle x^{1/2\times 7/3}=x^{7/6}\)

This problem relies on the key knowledge that \(\displaystyle \small x^\frac{a}{b}=\sqrt[b]{x^a}\) and that the multiplying terms with exponents requires adding the exponents. Therefore, we can rewrite the expression thusly:

\(\displaystyle \small a^\frac{3}{4}*a^\frac{3}{8}*a^\frac{5}{2}=a^{\frac{3}{4}+\frac{3}{8}+\frac{5}{2}}=a^{\frac{29}{8}}=\sqrt[8]{a^{29}}\)

Therefore, \(\displaystyle \small \sqrt[8]{a^{29}}\) is our final answer.

\(\displaystyle \bigg(\frac{27}{125}\bigg)^{\frac{1}{3}} = \frac{27^{\frac{1}{3}}}{125^{\frac{1}{3}}}=\frac{3}{5}\)

or

\(\displaystyle \bigg(\frac{27}{125}\bigg)^{\frac{1}{3}}= \sqrt[3]{\frac{27}{125}}=\frac{\sqrt[3]{27}}{\sqrt[3]{125}}=\frac{3}{5}\)

Keep in mind that when you are dividing exponents with the same base, you will want to subtract the exponent found in the denominator from the exponent found in the numerator.

To find the exponent for \(\displaystyle x\), subtract the denominator's exponent from the numerator's exponent.

\(\displaystyle \frac{1}{2}-(-\frac{3}{2})=2\)

To find the exponent for \(\displaystyle y\), subtract the denominator's exponent from the numerator's exponent.

\(\displaystyle 2-4=-2\)

Since the exponent is negative, you will want to put the \(\displaystyle y\) in the denominator in order to make it positive.

So then,

\(\displaystyle \frac{x^{\frac{1}{2}}y^2}{x^{\frac{-3}{2}}y^4}=\frac{x^2}{y^2}\)

When you have a number or value with a fractional exponent, 

\(\displaystyle x^{\frac{a}{b}}=\sqrt[b]{x^a}\)

or

\(\displaystyle x^{\frac{a}{b}}=(\sqrt[b]{x})^a\)

So then,

\(\displaystyle 9^{\frac{3}{2}}=\sqrt[2]{9^3}=\sqrt{729}=27\)

\(\displaystyle 9^{\frac{3}{2}}=(\sqrt9)^3=3^3=27\)

When you have a number or value with a fractional exponent, 

\(\displaystyle x^{\frac{a}{b}}=\sqrt[b]{x^a}\)

or

\(\displaystyle x^{\frac{a}{b}}=(\sqrt[b]{x})^a\)

So then,

\(\displaystyle 64^{\frac{4}{3}}=\sqrt[3]{64^4}=\sqrt[3]{2^{24}}=2^8=256\)

When exponents are raised to another exponent, you will need to multiply the exponents together.

When you have a number or value with a fractional exponent, 

\(\displaystyle x^{\frac{a}{b}}=\sqrt[b]{x^a}\)

or

\(\displaystyle x^{\frac{a}{b}}=(\sqrt[b]{x})^a\)

So,

\(\displaystyle 27^{\frac{5}{3}}=\sqrt[3]{27^5}=\sqrt[3]{3^{15}}=3^5=243\)

\(\displaystyle (t^3)^{\frac{5}{3}}=t^{3\times\frac{5}{3}}=t^{\frac{15}{3}}=t^5\)

\(\displaystyle (27t^3)^\frac{5}{3}=243t^5\)

An option to solve this is to split up the fraction.  Rewrite the fractional exponent as follows:

\(\displaystyle 25^{\frac{3}{2}} = (25^{\frac{1}{2}})^3\)

A value to its half power is the square root of that value.

\(\displaystyle 25^{\frac{1}{2}} = \sqrt{25}=5\)

Substitute this value back into \(\displaystyle (25^{\frac{1}{2}})^3\).

\(\displaystyle (25^{\frac{1}{2}})^3 = (5)^3 = 5\times 5\times 5 = 125\)