# Is this a special kind of exponential equation? How can I solve it?

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.

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 $$m=33$$.
• 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? Feb 16 '21 at 17:55
• 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. Feb 16 '21 at 19:00
• 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? Feb 17 '21 at 17:44
• 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 $n$.) Feb 17 '21 at 18:25