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

3. ## logs???

Originally Posted by JeWiSh
$\displaystyle 8x^\frac{-3}{4}=27$
$\displaystyle x^\frac{-3}{4}=\frac{27}{8}$
$\displaystyle -0.75logx=log\frac{27}{8}$
$\displaystyle logx=(log\frac{27}{8})/-0.75=D$ D is just for simplicity in the last line.
$\displaystyle 10^D=\frac{16}{81}$
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?

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

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

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

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

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

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