Exponents and Logs

Question

This is from an exam prep textbook - I've tried various methods including taking the log of all sides, the change of base theorem, and even inputting random numbers, but have had no luck.

Any help will be greatly appreciated, thanks in advance.

Answer 1

Let $a^{x}=b^{y}=(ab)^{xy}=Q$

By taking logarithm to the base Q we get that, $x\log_{Q}a=y\log_{Q}b=xy(\log_{Q}a+\log_{Q}b)=1$ as $\log ab=\log a+\log b$

We get that $\log_{Q}a=1/x,\log_{Q}b=1/y$

You can put the values of $\log_{Q}a,\log_{Q}b$ in $xy(\log_{Q}a+\log_{Q}b)=1$ to get

$xy(1/x + 1/y) = 1$ from which we get that $x + y = 1$

EDIT - A similar question was asked here. This link also has 3 different solutions to the problem one of which is same as mine - Prove that if $a^x = b^y = (ab)^{xy}$, then $x + y = 1$