The infinitely many solutions of $e^{2z-3} = 5$

I have $$e^{2z-3} = 5$$

I'm trying to find ALL the complex numbers that satisfy this but can't get my head around how to do it?

I can solve for z, this gave me $z = {(\ln5 +3)}/2$

But for complex numbers asking for multiple answers this doesn't seem the logical solution?

Any advice would be brilliant

$$e^{2z-3} = 5$$ Letting $z=x+i\,y:$ $$e^{2x-3}e^{i\,2y}=5$$ Taking the complex logarithm: $$2x-3+i(2y+2m\pi)=\ln5+i(2n\pi)$$ Equating the real and imaginary parts: $$x=\frac{3+\ln5}2\; \text{ and }\; y=k\pi$$ I.e., $$z=\frac{3+\ln5}2+i\,k\pi.$$

(Since every step above is reversible, no extraneous solution has been created.)

Hint: You're almost there. Remember that $f(z) = e^z$ is periodic with period $2\pi i$ and is 1-1 on each horizontal strip $S_a \equiv\{z : a \leq \operatorname{Im} z < a + 2\pi\}$, so if you find one solution you can find all the others.

Your function is a slight modification of this--the factor of $2$ must be taken into account.