Question 398745
{{{(25/x^(-4))^(3/2)}}}
First let's work within the parentheses. In general {{{a^(-n) = 1/a^n}}} and {{{1/a^(-n) = a^n}}}. Using the second pattern, the expression within the parentheses becomes:
{{{(25x^4)^(3/2)}}}
Now we can use a rule for exponents, {{{(ab)^n = a^n*b^n}}}, to raise {{{25x^4}}} to the 3/2 power:
{{{25^(3/2)*(x^4)^(3/2)}}}
Simplifying the last part of the expression is simple. The rule for exponents when raising a power to a power is to multiply the exponents:
{{{25^(3/2)*x^(4*(3/2))}}}
{{{25^(3/2)*x^6}}}
Simplifying the first part can be difficult for those who do not yet understand fractional exponents. I find that it can help if you factor the exponent in a special way:<ol><li>If the exponent is negative, factor out -1.</li><li>If the exponent is fractional and the numerator is not a 1, then factor out the numerator.</li></ol>
The exponent on the 25 is not negative. But it is fractional and its denominator is not a 1. So we will factor out the 3:
{{{25^(3*(1/2))*x^6}}}
Each of the factors tells us an operation to perform. The 3 tells us that we will be cubing. And the 1/2, if you remember what fractional exponents mean, tells us that we will be finding a square root. And since multiplication is Commutative, we can do these operations <i>in any order we choose.</i> Cubing 25 doesn't look very appealing while finding a square root of 25 looks very easy. So we will start with the square root:
{{{25^((1/2)*3)*x^6}}}
{{{(25^(1/2))^3*x^6}}}
{{{(5)^3*x^6}}}
{{{125x^6}}}<br>
Note: If we had chosen to cube 25 first and then find a square root, we would still have gotten 125 -- just not as easily.