Question 352453
I assume the expression is:
{{{root(4, x)*root(3, 6x)}}}
If this is correct, the first radical is read: "the fourth root of x" and the second radical is read: "the third root of 6x" or "the cube root of 6x". The 4 and the 3 are <b>not</b> exponents (although, as you'll see shortly, they do become denominators of fractional exponents).<br>
Another word for "rational" (in Math) is "fractional". So the problem is saying: "Use fractional exponents to ..." This means we need to know how to change radicals into fractional exponents (and vice versa). The connection between radicals and exponents is:
{{{q^(a/b) = root(b, q^a) = (root(b, q))^a}}}
(Note: square roots are "second roots". IOW, {{{sqrt(x)}}} is the same thing as {{{root(2, x)}}}.)<br>
So your expression, written with fractional exponents instead of radicals is:
{{{x^(1/4)*(6x)^(1/3)}}}
(Note: Are you sure the 6 is in the problem? It makes the problem much harder.) Using the the rule for exponents, {{{(a*b)^q = a^q*b^q}}}, on the 6x part of the above expression we get:
{{{x^(1/4)*6^(1/3)x^(1/3)}}}
To multiply the "x" parts of the above we will use a rule for exponents, {{{q^a*q^b = q^(a+b)}}}, which says we should add the exponents:
{{{6^(1/3)x^(((1/3) + (1/4)))}}}
To add the exponents, which are fractions, we must, of course, have common denominators:
{{{6^(1/3)x^(((4/12) + (3/12)))}}}
{{{6^(1/3)x^(7/12)}}}
To combine the 6 part and the "x" part of the above we will use the {{{(a*b)^q = a^q*b^q}}} rule again (in the opposite direction this time). Once again we need a common denominator:
{{{6^(4/12)x^(7/12)}}}
Factoring out a 1/12 in each exponent we get:
{{{6^((4*(1/12)))x^((7*(1/12)))}}}
Now we can use the {{{(a*b)^q = a^q*b^q}}} rule (from right to left this time):
{{{(6^4x^7)^(1/12)}}}
Now, since 1/12 as an exponent means 12th root, we can rewrite the above as the radical:
{{{root(12, 6^4*x^7)}}}
The only thing left is to multiply out {{{6^4}}}. I'll leave that to you and your calculator. (It should work out to be a number somewhere near 1300.)