Solving natural log equation for variable

Question

I am very confused on how to solve this equation for $k$; I've tried using $ln(x)$, but my math isn't coming out; how can this be solved?

$$3\gamma e^{\gamma k-15000}=-2(1-\gamma)e^{-15000-2k(1-\gamma)}$$

Answer 1

The correct expression after applying $\log$ is the following:

$$\log(3)+\log(\gamma)+[\gamma k-15000]=\log(2)+ \log(\gamma -1) + [-15000-2k(1-\gamma)]$$

Can you take it from here?

Answer 2

Alternatively: $$3\gamma e^{\gamma k-15000}=-2(1-\gamma)e^{-15000-2k(1-\gamma)} \Rightarrow$$ $$e^{\gamma k-15000+15000+2k(1-\gamma)}=\frac{2(\gamma-1)}{3\gamma} \Rightarrow$$ $$e^{2k-k\gamma}=\frac{2(\gamma-1)}{3\gamma} \stackrel{\ln}\Rightarrow$$ $$2k-k\gamma=\ln{\frac{2(\gamma-1)}{3\gamma}}$$ $$k=\frac{1}{2-\gamma}\cdot \ln{\frac{2(\gamma-1)}{3\gamma}}.$$