# Math Help - Radical Equations

1. ## Radical Equations

I found the following question in my Math B book.
It is a radical equation problem. I played with this question for 20 minutes and got nowhere.

Here it is:

8x^(-3/4) = 27

Solve for x.

The answer is x = 16/81. How do I get this answer?

2. Originally Posted by magentarita
I found the following question in my Math B book.
It is a radical equation problem. I played with this question for 20 minutes and got nowhere.

Here it is:

8x^(-3/4) = 27

Solve for x.

The answer is x = 16/81. How do I get this answer?
$8x^\frac{-3}{4}=27$
$x^\frac{-3}{4}=\frac{27}{8}$
$-0.75logx=log\frac{27}{8}$
$logx=(log\frac{27}{8})/-0.75=D$ D is just for simplicity in the last line.
$10^D=\frac{16}{81}$

3. ## logs???

Originally Posted by JeWiSh
$8x^\frac{-3}{4}=27$
$x^\frac{-3}{4}=\frac{27}{8}$
$-0.75logx=log\frac{27}{8}$
$logx=(log\frac{27}{8})/-0.75=D$ D is just for simplicity in the last line.
$10^D=\frac{16}{81}$
I know about applying logs but this question is located in the radical equations chapter. How do you find x using radical equations?

4. Originally Posted by magentarita
I found the following question in my Math B book.
It is a radical equation problem. I played with this question for 20 minutes and got nowhere.

Here it is:

8x^(-3/4) = 27

Solve for x.

The answer is x = 16/81. How do I get this answer?

$8x^{-\frac{3}{4}}=27$

$x^{-\frac{3}{4}}=\frac{27}{8}$

$
x=(\frac{27}{8})^{-\frac{4}{3}}
$

5. Just to elaborate, we're making use of the fact that:

$\left(\displaystyle x^{ \displaystyle \frac{a}{b}}\right)^{ \displaystyle \frac{b}{a}} = \displaystyle x$

So for your question, mathaddict raised both sides to the power of $-\tfrac{4}{3}$ .