# Question 7fed6

Oct 15, 2015

You may write radicals as fractional powers.

#### Explanation:

$\sqrt[5]{{a}^{4}} = {a}^{\frac{4}{5}}$

$\sqrt[3]{{a}^{2}} = {a}^{\frac{2}{3}}$

In multiplication, you add the powers, so:

root 5 (a^4)*root 3 (a^2)=a^(4/5)*a^(2/3)=a^((4/5+2/3)=

${a}^{\left(\frac{12}{15} + \frac{10}{15}\right)} = {a}^{\frac{22}{15}}$ so $n = \frac{22}{15}$

Oct 15, 2015

$n = \frac{22}{15}$

#### Explanation:

Here's an alternative approach

$\sqrt[5]{{a}^{4}} \cdot \sqrt[3]{{a}^{2}} = {a}^{n}$

Once again, the first thing to do is rewrite your radicals as exponents by using

$\textcolor{b l u e}{\sqrt[y]{{a}^{x}} = {a}^{\frac{x}{y}}}$

$\sqrt[5]{{a}^{4}} = {a}^{\frac{4}{5}} \text{ }$ and $\text{ } \sqrt[3]{{a}^{2}} = {a}^{\frac{2}{3}}$

Now focus on the exponents. Find their common denominator, which in this csae is $15$. You can write

$\frac{4}{5} \cdot \frac{3}{3} = \frac{12}{15} \text{ }$ and $\text{ } \frac{2}{3} \cdot \frac{5}{5} = \frac{10}{15}$

Now the equation becomes

${a}^{\frac{12}{15}} \cdot {a}^{\frac{10}{15}} = {a}^{n}$

If you want to play aroun with the exponents a bit, you can convert back to radical form

${a}^{\frac{12}{15}} = \sqrt[15]{{a}^{12}} \text{ }$ and $\text{ } {a}^{\frac{10}{15}} = \sqrt[15]{{a}^{10}}$

Now you have

${a}^{\frac{12}{15}} \cdot {a}^{\frac{10}{15}} = \sqrt[15]{{a}^{12}} \cdot \sqrt[15]{{a}^{10}} = {a}^{n}$

This is equivalent to

$\sqrt[15]{{a}^{12} \cdot {a}^{10}} = {a}^{n}$

Once again,

color(blue)(a^x * a^y = a^(x+y)#

so you get

$\sqrt[15]{{a}^{12 + 10}} = {a}^{n}$

$\sqrt[15]{{a}^{22}} = {a}^{n}$

Finally, convert back to exponent form to get

${a}^{\frac{22}{15}} = {a}^{n} \implies n = \textcolor{g r e e n}{\frac{22}{15}}$