Thread: How do I solve for this exponent using logs?

1. How do I solve for this exponent using logs?

(4.0)(8.0)^z = (7.0)(3.0)^(z-1)

I have to solve for z. I'm reviewing basic logarithm rules but I can't remember how to do this at all.

Can someone provide some insight? Thanks!

2. Hello,
Originally Posted by Jeavus
(4.0)(8.0)^z = (7.0)(3.0)^(z-1)

I have to solve for z. I'm reviewing basic logarithm rules but I can't remember how to do this at all.

Can someone provide some insight? Thanks!
Just use the following rules :

$\displaystyle \log(ab)=\log(a)+\log(b)$

$\displaystyle \log(a^b)=b \log(a)$

3. Do I have to take the log of just the (8.0)^z and the (3.0)^(z-1)? Or the entire term itself? That's where I'm having trouble...

4. Originally Posted by Jeavus
Do I have to take the log of just the (8.0)^z and the (3.0)^(z-1)? Or the entire term itself? That's where I'm having trouble...
$\displaystyle a=b \Longleftrightarrow \log(a)=\log(b)$ (a and b positive)

So $\displaystyle \log(4 \cdot 8^z)=\log(7 \cdot 3^{z-1})$

5. Thanks! I got it!