Question 257399
When working with fractional exponents you need to know about the properties of exponents:<ol><li>{{{a^p * a^q = a^(p+q)}}}</li><li>{{{a^p / a^q = a^(p-q)}}}</li><li>{{{(a^p)^q = a^(p*q)}}}</li><li>{{{(ab)^p = a^p*b^p}}}</li><li>{{{a^(-p) = 1/a^p}}}</li><li>{{{a^(p/q) = root(q, a^p) = (root(q, a))^p}}}</li></ol>
Let's see how these properties help us simplify your expressions:
1. {{{(-16)^(1/4)}}}
By property #6
{{{(-16)^(1/4) = root(4, (-16)^1) = root(4, -16)}}}
This expression represents the 4th root of -16. In other words the number which, when raised to the 4th power, results in -16. There are no such numbers in the set of Real numbers. Whenever we raise a Real number to the 4th power we always get a positive result. If you have never heard of Complex Numbers then we can go no further so skip to #2.<br>
If you know about Complex Numbers, Complex Numbers in Polar form and finding roots of Complex Numbers then we can come up with an answer. Writing {{{(-16)^(1/4)}}} as a Complex Number in Polar form we get:
{{{(16(cos(pi)+i*sin(pi)))^(1/4)}}}
Using deMoivre's (spelling?) Theorem this is equal to:
{{{16^(1/4)(cos((1/4)pi)+i*sin((1/4)pi))}}}
Which simplifies as follows:
{{{(2)(sqrt(2)/2+i*sqrt(2)/2)}}}
{{{sqrt(2)+i*sqrt(2)))}}} This is the primary 4th root. The other 3 are:
{{{sqrt(2)-i*sqrt(2)))}}}
{{{-sqrt(2)+i*sqrt(2)))}}}
{{{-sqrt(2)-i*sqrt(2)))}}}
<br>
2. {{{25^(-3/2)}}}
By property #5 above this is equal to:
{{{1/25^(3/2)}}}
By property #6 the denominator can be written as {{{sqrt(25^3)}}} or {{{(sqrt(25))^3}}}. Since the second form looks easier (after all we do know what the square root of 25 is) we will use that form:
{{{1/(sqrt(25))^3 = 1/(5)^3 = 1/125}}}<br>
3. {{{(-27x^9)^(1/3)}}}
Using property #4 above we get:
{{{(-27)^(1/3)*(x^9)^(1/3)}}}
Using property #6 on the first part and property #3 on the second part we get:
{{{root(3, -27)*x^(9*(1/3))}}}
which simplifies to:
{{{-3x^3}}}