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$e^{i\pi} = e^{z\ln2} + e^{z\ln3}$

How can I deal with this? I want to solve for z. Does this help?

$e^{z\ln2} + e^{z\ln3} = e^{z\ln2}(1 + e^{z(ln3-ln2)})$

If I write out z=x+iy then the expression becomes

$-1 = e^{x\ln2}e^{iy\ln2}+e^{x\ln3}e^{iy\ln3}$

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    $\begingroup$ Not hard to simplify it to $-1=2^z + 3^z$, but from there, I don't know. Edit: just to show that simplification: $$e^{i\pi} = e^{z\ln2} + e^{z\ln3}$$ $$-1 = (e^{\ln2})^{z} + (e^{\ln3})^{z}$$ $$-1= 2^{z} + 3^{z}$$ $\endgroup$ – Sultan of Swing Dec 20 '14 at 1:30
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    $\begingroup$ Yeah, it looks nontrivial :) Just remember $e^{i\pi} = -1$ $\endgroup$ – Sultan of Swing Dec 20 '14 at 1:36
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    $\begingroup$ You could use Euler's formula to identify real and imaginary parts, but I still get stuck. $$-1 = 2^x(\cos y\ln2 + i\sin y\ln 2)+3^x(\cos y\ln3 + i\sin y\ln3)$$ Which means: $$-1 = 2^x\cos y\ln2 + 3^x\cos y\ln3 \qquad 0 = 2^x\sin y\ln2 + 3^x\sin y\ln3.$$ I don't see a continuation... $\endgroup$ – dietervdf Dec 20 '14 at 2:17
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    $\begingroup$ @dietervdf can I just suggest that you write $\sin(y \ln 3)$ instead of $\sin y \ln 3$ in your comment? This ambiguity messed me up earlier :) $\endgroup$ – Sultan of Swing Dec 20 '14 at 4:25
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    $\begingroup$ My answer is hardly exemplary...but I really appreciate the sentiments (the bounty). I don't feel I am worthy of the number of upvotes I got on my answer, let alone an additional bounty. If you check the edit history of my answer, you'll see I did some work to try and solve it, but that work was incorrect as I made the mistake of confusing $\sin y \ln 3$ with $\ln 3 (sin y)$, so everything turned out to be incorrect. Even if that work was correct, that wouldn't have been that fantastic of an answer :) $\endgroup$ – Sultan of Swing Dec 25 '14 at 20:08
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To the best of my knowledge, you can't solve for $z$ analytically. As in my comment above, the expression simplifies to $-1=2^z + 3^z$, but there isn't a way to find $z$ explicitly. Even separating into real and imaginary parts yields a system of nonlinear equations. So to answer: is it solvable? No, it is not. So you'd have to use numerical methods to solve $0=2^z + 3^z -1$. But in terms of interpreting it, I hope someone can provide more insight.

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Write $log(2)=A+2\pi i n$ and $log(3)=B + 2\pi i m$ where $A$ and $B$ are the real logs of $2$ and $3$.

Write $z=x+iy$, write out the real and imaginary parts of your equation and you'll get two equations in the four unknowns (two real, two integers) $x,y,m,n$, and it won't be hard to characterize the solutions.

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  • $\begingroup$ Can you please elaborate how we make it two equations? $\endgroup$ – Niklas R. Dec 20 '14 at 2:10
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    $\begingroup$ @NiklasRtz: The real part of the left-hand side is equal to the real part of the right-hand side and the imaginary part of the left-hand side is equal to the imaginary part of the right-hand side. $\endgroup$ – WillO Dec 20 '14 at 3:07
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    $\begingroup$ The problem is that you end up solving two nonlinear equations in two (or four, if you rewrite the logarithms) unknowns, which I see no method of solving. $\endgroup$ – Sultan of Swing Dec 20 '14 at 3:13

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