# Thread: Raised by an unknown quantity

1. ## Raised by an unknown quantity

Hi! Here I'm again!

If $\displaystyle 5^{3a} = 64$ , so $\displaystyle 5^{-a}$ is?

I've tried change the base, ($\displaystyle 64 = 2^6$), for example... But I coudn't get a result, because I don't know what to do with the $\displaystyle 5^3 = 125$ and the $\displaystyle 5^a$ , or how to transform they on $\displaystyle 2$

>>>

$\displaystyle 5^{3a} = 64$

$\displaystyle 5^3\times5^a = 2^6$

This is the way?
How can I do this?!

2. Originally Posted by PxEckx
Hi! Here I'm again!

If $\displaystyle 5^{3a} = 64$ , so $\displaystyle 5^{-a}$ is?

I've tried change the base, ($\displaystyle 64 = 2^6$), for example... But I coudn't get a result, because I don't know what to do with the $\displaystyle 5^3 = 125$ and the $\displaystyle 5^a$ , or how to transform they on $\displaystyle 2$

>>>

$\displaystyle 5^{3a} = 64$

$\displaystyle 5^3\times5^a = 2^6$

This is the way?
How can I do this?!
Hi PxEckx,

$\displaystyle \displaystyle \left(5^{-a}\right)^{-3}=5^{3a}=64$

$\displaystyle \displaystyle \left(\left(5^{-a}\right)^{-3}\right)^{-\frac{1}{3}}=\left(64\right)^{-\frac{1}{3}}$

$\displaystyle \displaystyle 5^{-a}=\left(64\right)^{-\frac{1}{3}}=\text{?}$

3. I forgot this 'peculiarity'! Perfect! Rise both sides by the same number!!!

So:::

$\displaystyle (5^{-a})^{-3} = 5^{3a} = 64$

$\displaystyle (5^{3a})^-\displaystyle^\frac{1}{3} = (64)^-^\frac{1}{3}$

$\displaystyle 5^{-a} = \displaystyle\frac{1}{\sqrt[3]{64} }$

$\displaystyle 5^{-a} = \displaystyle\frac{1}{4}$

Thanx!!!