1. ## Indices, logarithms question

hey there
been stuck on this one for ages. can't really get a grasp on it

Given: $\displaystyle 3^x = 4^y = 12^z$ , show that $\displaystyle z = \frac{xy}{x+y}$

2. ## Re: Indices, logarithms question

Originally Posted by walleye
hey there
been stuck on this one for ages. can't really get a grasp on it

Given: $\displaystyle 3^x = 4^y = 12^z$ , show that $\displaystyle z = \frac{xy}{x+y}$
Here's one way you can do it :
We know that $\displaystyle 3^x = 4^y = 12^z$
now taking logarithms, this equation becomes $\displaystyle \implies x\log 3 = y\log 4 = z\log 12$
we also write:$\displaystyle x\log 3 = y\log 4 = z\log 12 = k$
now $\displaystyle \log 3=\frac{k}{x}$ and $\displaystyle \log 4=\frac{k}{y}$

Also,$\displaystyle z\log 12 = k \implies z\log(4.3)=k$
by applying the log properties:$\displaystyle \implies z\log 4 + z\log 3 = k \implies z(\log 4 +\log 3)=k$
putting in the values of $\displaystyle \log 4$ and $\displaystyle \log 3$, $\displaystyle \implies$ $\displaystyle zk(\frac{1}{x}+\frac{1}{y})=k$
$\displaystyle k$ gets cancelled,and simplifying gives us: $\displaystyle z(\frac{x+y}{xy})=1$
which gives $\displaystyle z=\frac{xy}{x+y}$