Question #7fed6

`root 5 (a^4)=a^(4/5)`

`root 3 (a^2)=a^(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^((12/15+10/15))=a^(22/15)` so `n=22/15`

Here's an alternative approach

`root(5)(a^4) * root(3)(a^2) = a^n`

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

`color(blue)(root(y)(a^x) = a^(x/y))`

In your case, you have

`root(5)(a^4) = a^(4/5)" "` and `" "root(3)(a^2) = a^(2/3)`

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

`4/5 * 3/3 = (12)/15" "` and `" "2/3 * 5/5 = 10/15`

Now the equation becomes

`a^(12/15) * a^(10/15) = a^n`

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

`a^(12/15) = root(15)(a^12)" "` and `" "a^(10/15) = root(15)(a^(10))`

Now you have

`a^(12/15) * a^(10/15) = root(15)(a^12) * root(15)(a^(10)) = a^n`

This is equivalent to

`root(15)( a^12 * a^10) = a^n`

Once again,

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

so you get

`root(15)(a^(12 + 10)) = a^n`

`root(15)(a^(22)) = a^n`

Finally, convert back to exponent form to get

`a^(22/15) = a^n implies n = color(green)(22/15)`