simplifying equation with logs I have the following equation:

I would like to solve this for Ze. I have found the same equation expressed in terms of Ze in another paper:

I can't get my head around how this works. This is my attempt:

However, there is something about the last step that doesn't seem right to me, even though the answers match. 
Can anyone explain the correct process for simplifying the first equation in terms of Ze? 
 A: Be careful:


*

*$e^{a+b}$ is not the same as $e^a+e^b$ (it's actually $e^a.e^b$).

*When it comes to the final step, one only needs to note that $e^{-a}=\frac{1}{e^a}$.


The full derivation might look like this:
$$\ln\left(\frac{z_0}{z_e}\right) = b_1Re^{0.25}+b_2$$
Get rid of the logarithm:
$$\frac{z_0}{z_e} = \exp(b_1Re^{0.25}+b_2)$$
Multiply by $z_e$ and divide by the exponential:
$$z_e = z_0\frac{1}{\exp(b_1Re^{0.25}+b_2)}$$
Use the fact that $\exp{-a}=\frac{1}{\exp{a}}$:
$$z_e = z_0\exp(-(b_1Re^{0.25}+b_2))$$
A: You made a mistake in the first step. Here's how it should work:
$$\begin{align*}
\ln\left(\frac{z_0}{z_e}\right) &= b_1Re^{0.25}+b_2 \\
\exp\left(  \ln\left(\frac{z_0}{z_e}\right) \right) &= \exp\left( b_1Re^{0.25}+b_2 \right) \\
\frac{z_0}{z_e} &= \exp\left( b_1Re^{0.25} + b_2 \right) \\
\left(\frac{z_0}{z_e}\right)^{-1} &=\left( \exp\left( b_1Re^{0.25} + b_2 \right) \right)^{-1} \\
\frac{z_e}{z_0} &= \exp\left(-1\left( b_1Re^{0.25} + b_2 \right) \right) \\
\frac{z_e}{z_0} &= \exp\left(-b_1Re^{0.25} - b_2 \right) \\
z_e &= z_0\exp\left(-b_1Re^{0.25} - b_2 \right) \\
\end{align*}$$
Note that since $x^a \cdot x^b = x^{a+b}$ and $(x^a)^b=x^{ab}$, we have the following identities:

  
*
  
*$\exp(a+b)=\exp(a) \cdot \exp(b)$
  
*$\left(\exp(a)\right)^b=\exp(ab)$
  

