# Thread: high school algebra problem

1. ## high school algebra problem

Having some trouble with this one.

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

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

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

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

Having some trouble with this one.

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

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

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

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

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

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

$\left(\frac{2}{3}\right)^3 = \frac{8}{27}$

Having some trouble with this one.

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

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

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

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 $\left( \frac ab \right)^n = \frac {a^n}{b^n}$ and $\left( \frac ab \right)^{-n} = \frac 1{\left( \frac ab \right)^n} = \left( \frac ba \right)^n$

So, then

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

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

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

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

$= \frac 8{27}$

4. thanks for that guys
the 3/4 was throwing me off