Question #7fed6

2 Answers
Oct 15, 2015

Answer:

You may write radicals as fractional powers.

Explanation:

#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#

Oct 15, 2015

Answer:

#n = 22/15#

Explanation:

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)#