# high school algebra problem

• Mar 14th 2010, 05:36 PM
high school algebra problem
Having some trouble with this one.

My understanding so far is that if $\displaystyle x^-1$ then it equals $\displaystyle 1/x$. Also that $\displaystyle x^1/2$ equals $\displaystyle sqrt x$.

The problem I am trying to simplify is $\displaystyle (81/16)^-3/4$.

I have simplified it to $\displaystyle 4sqrt (16/81)^3$

Can someone help me in the right direction please?

Not sure what the code for square root is in the math writing program either :S
• Mar 14th 2010, 05:43 PM
skeeter
Quote:

Having some trouble with this one.

My understanding so far is that if $\displaystyle x^-1$ then it equals $\displaystyle 1/x$. Also that $\displaystyle x^1/2$ equals $\displaystyle sqrt x$.

The problem I am trying to simplify is $\displaystyle (81/16)^-3/4$.

I have simplified it to $\displaystyle 4sqrt (16/81)^3$

Can someone help me in the right direction please?

Not sure what the code for square root is in the math writing program either :S

$\displaystyle \left(\frac{81}{16}\right)^{-\frac{3}{4}} =$

$\displaystyle \left(\frac{16}{81}\right)^{\frac{3}{4}} =$

$\displaystyle \left[\left(\frac{2^4}{3^4}\right)^{\frac{1}{4}}\right]^3 =$

$\displaystyle \left(\frac{2}{3}\right)^3 = \frac{8}{27}$
• Mar 14th 2010, 05:44 PM
Jhevon
Quote:

Having some trouble with this one.

My understanding so far is that if $\displaystyle x^-1$ then it equals $\displaystyle 1/x$. Also that $\displaystyle x^1/2$ equals $\displaystyle sqrt x$.

The problem I am trying to simplify is $\displaystyle (81/16)^-3/4$.

I have simplified it to $\displaystyle 4sqrt (16/81)^3$

Can someone help me in the right direction please?

Not sure what the code for square root is in the math writing program either :S

see here for instructions on using LaTeX, the markup language built into the forum through which you can get pretty math symbols

Now, note that $\displaystyle \left( \frac ab \right)^n = \frac {a^n}{b^n}$ and $\displaystyle \left( \frac ab \right)^{-n} = \frac 1{\left( \frac ab \right)^n} = \left( \frac ba \right)^n$

So, then

$\displaystyle \left( \frac {81}{16} \right)^{-3/4} = \left( \frac {16}{81} \right)^{3/4}$

$\displaystyle = \frac {16^{3/4}}{81^{3/4}}$

$\displaystyle = \frac {( \sqrt[4]{16})^3}{(\sqrt[4]{81})^3}$

$\displaystyle = \frac {2^3}{3^3}$

$\displaystyle = \frac 8{27}$
• Mar 14th 2010, 05:53 PM