(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!
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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 : $\displaystyle \log(ab)=\log(a)+\log(b)$ $\displaystyle \log(a^b)=b \log(a)$
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...
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 $\displaystyle \log(4 \cdot 8^z)=\log(7 \cdot 3^{z-1})$
Thanks! I got it!
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