<p>[url=<a href="http://img62931.pictiger.com/images/13744703/%5Dsat%5B/url">http://img62931.pictiger.com/images/13744703/]sat[/url</a>]</p>
<p>btw thats a 3 for those of you who cant see it(number outside the square root in the denominator of the original problem)</p>
<p>How do you go about solving that? Simplifying? How?</p>
<p>The square root of x cab be written as x^(1/2), where the denominator of the fraction is the value of the radical (square root-2, cube root-3, etc) and the numerator is the power of x inside, in this case its one. So we have x to the 1/2</p>
<p>So rewrite the problem using this fraction notation instead of radicals</p>
<p>it becomes x^(5/6) over x^(2/3). Make the 2/3 a 4/6 so the fractions have the same denominator. So its (x^(5/6))/(x^(4/6)). Now you know the law of exponents, when two numbers of the same base are divided by each other, you subtract the exponent on the denominator from the exponent on the numerator, and the result will the the base to that calculated exponent. Basically, (5/6) -(4/6) which is (1/6). So we have x^(1/6).</p>
<p>Now, rewrite it as a radical. If x to the (1/2) means the square root of x to the first power, then x^(1/6) means the sixth root of x to the first power. Your answer is D.</p>
<p>Thanks I Get It Wooo</p>
<p>Thanks I Was Gonna Put It Into Fractions But I Got Stuck After That ..ugh </p>
<p>Thanks</p>