Question 249706
the 8th root of {{{16x^4*y^12}}}<br>
First we need to know what this means. "the 8th root of {{{16x^4*y^12}}} means {{{root(8, 16x^4*y^12)}}}
Second, it helps if we understand how to write this using fractional exponents.
{{{root(8, 16x^4*y^12) = (16x^4*y^12)^(1/8)}}}
Next, we should recognize that {{{16x^4*y^12)}}} is a single term (with many factors). Since it is a single term, the following rule for exponents applies:
{{{(a*b*c)^d = a^d*b^d*c^d}}}
(This looks similar to the Distributive Property <i>but it is <b>not</b> the Distributive Property.</i>) We can use this to raise each factor to the 1/8th power:
{{{(16x^4 y^12)^(1/8) = 16^(1/8)*(x^4)^(1/8)*(y^12)^(1/8)}}}
Since {{{16 = 2^4}}} we can rewrite this as:
{{{(2^4)^(1/8)*(x^4)^(1/8)*(y^12)^(1/8)}}}
Now we can use another rule for exponents, {{{(a^p)^q = a^(p*q)}}}, to simplify the expression:
{{{2^(4*(1/8))*x^(4*(1/8))*y^(12*(1/8))}}}
which simplifies to:
{{{2^(1/2)*x^(1/2)*y^(3/2)}}}
(Can you now see why we changed {{{16}}} into {{{2^4}}}?)
Since all the exponents have the same denominator they all represent the same root: square root. We can factor out a 1/2 from each exponent using the earlier property (in the other direction):
{{{(2xy^3)^(1/2)}}}
To finish this we will write it will a radical:
{{{sqrt(2xy^3)}}}
The only thing left here is to simplify, if possible. Simplifying square roots involves finding perfect square factors, if any. We do have a perfect square factor, {{{y^2}}}, which we can factor out:
{{{sqrt(2xy^3) = sqrt(y^2*2xy) = sqrt(y^2)*sqrt(2xy) = abs(y)sqrt(2xy)}}}
(Note that we can't say {{{sqrt(y^2) = y}}} because square roots are positive but y does not have to be positive. So we use {{{abs(y)}}} to ensure that we end up with a positive number.)