# How do you simplify the fifth root of t (or t^(1/5)) multiplied by radical (16t^5)?

Mar 22, 2015

I assume that "radical" here means square root.

${t}^{\frac{1}{5}} \cdot \sqrt{16 {t}^{5}} = {t}^{\frac{1}{5}} \sqrt{16} {\sqrt{t}}^{5} = 4 {t}^{\frac{1}{5}} \sqrt{{t}^{5}}$

Now as a general definition: $\sqrt[n]{{x}^{m}} = {x}^{\frac{m}{n}}$. So $\sqrt{{t}^{5}} = {t}^{\frac{5}{2}}$

The expression now becomes:

${t}^{\frac{1}{5}} \cdot \sqrt{16 {t}^{5}} = 4 {t}^{\frac{1}{5}} \sqrt{{t}^{5}} = 4 {t}^{\frac{1}{5}} {t}^{\frac{5}{2}}$

An important property of exponents assures us that:
${x}^{n} {x}^{m} = {x}^{n + m}$.

So we can simplify ${t}^{\frac{1}{5}} {t}^{\frac{5}{2}}$ by adding $\frac{1}{5} + \frac{5}{2} = \frac{27}{10}$.

${t}^{\frac{1}{5}} \cdot \sqrt{16 {t}^{5}} = 4 {t}^{\frac{1}{5}} {t}^{\frac{5}{2}} = 4 {t}^{\frac{27}{10}} = 4 \sqrt[10]{{t}^{27}}$.