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I like looking through random equations online, and can't figure out what kind of equation this is:

$$ 2^x + 2^y = 2^z$$

Seems like a multivariate exponential equation, but the results on the internet are all basic. I'm trying to solve 'Find all x, y, z such that $$ x+y+z = 100 $$ $$ 2^x + 2^y = 2^z$$'

Would appreciate help on this question I thought of while studying exponential equations. I tried using logarithms and laws of exponents but they didn't really help.

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Assuming we are looking for integer solutions, let $m$ be the smallest of $x,y,z$. Then $2^{m+1}$ would divide the other two terms and therefore the $2^m$ term as well (which is absurd) unless two of $x,y,z$ equal $m$.

We then have $2^m + 2^m = 2^{m+1}$ and so $m+m+(m+1)=100$ and $m=33$.

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  • $\begingroup$ Thanks this helped me work out other problems where the number of terms on the LHS matches the base! But what if it were $ 2^x + 2^y + 2^z = 2^b $? With $ x+y+z+b = 100$? Would this still be solvable somehow? $\endgroup$
    – porusmaria
    Feb 16 '21 at 17:55
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    $\begingroup$ Glad to have helped. Yes, this equation can be solved by the same method. As before we know that two of $x,y,z,b$ equal the minimum power $m$. Clearly $b$ cannot be the minimum so suppose $x=y=m$. Then $2^{m+1}+2^z=2^b$ and we have an equation that we have already solved. $\endgroup$
    – S. Dolan
    Feb 16 '21 at 19:00
  • $\begingroup$ Insightful as always! Thanks! Can this method be generalized to any equation of the form $ n^{x_{1}} + n^{x_{2}} + ... + n^{x_{k-1}}= n^{x_{k}} $ for any integer n and k? Or are there certain restrictions? $\endgroup$
    – porusmaria
    Feb 17 '21 at 17:44
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    $\begingroup$ The method can certainly be generalised. If $m$ is the minimum power, then the number of $x_i, 1\le i\le k-1$ which are equal to $m$ must be a multiple of $n$ and the problem will have been massively reduced ( unless $k$ is huge compared to $n$.) $\endgroup$
    – S. Dolan
    Feb 17 '21 at 18:25
  • $\begingroup$ Thanks for your answer! Please allow me some time to think through this generalization; I'm afraid I'm not too mathematically savvy (yet). $\endgroup$
    – porusmaria
    Feb 21 '21 at 16:02

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