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Suppose $A$ and $B$ are positive real numbers for which $\log_AB=\log_BA$. If neither $A$ nor $B$ is $1$, and if $A\neq B$, find the value of $AB$.

So I use the change of base theorem getting $$\frac{\log B}{\log A}=\frac{\log A}{\log B}$$ I then cross multiply getting $$\left(\log A\right)^2=\left(\log B\right)^2$$ which simplifies to $$\log A=\log B$$ It seems that this is a dead end, as I see no other solution other than $A=B$.

I could also go on to have $$\frac{\log A}{\log 5+\log2}=\frac{\log B}{\log 5+\log2}$$ which would give me $$\log A(\log 5+\log2)=\log B(\log 5+\log2)$$ but sadly, I don't know how to multiply logs so I'm stuck this way.

Going literally by the log definition gives me $$B=A^{\log_BA}$$ and doesn't get anywhere. Help would be appreciated!

Also, if you are nice, could you also help me on this($N$'s base-5 and base-6 representations, treated as base-10, yield sum $S$. For which $N$ are $S$'s rightmost two digits the same as $2N$'s?) problem?

Thanks!

Max0815

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You can avoid going to a third base. Just use that $$ \log_BA=\frac{1}{\log_AB} $$ (which is legal because $B\ne1$) so your equation yields $(\log_AB)^2=1$, hence $$ \log_AB=1 \qquad\text{or}\qquad \log_AB=-1 $$ The former implies $A=B$, so it has to be discarded. Hence $B=A^{-1}$.

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  • $\begingroup$ Oh, a generalization of it, thanks! $\endgroup$ – Max0815 Feb 14 '19 at 23:38
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$x^{2}=y^{2}$ does not imply $x=y$. It implies $x=y$ or $x =-y$. Hence $\log\, A =\pm \log, B$. Since $A \neq B$ we get $\log\, A =-\log, B$ which van be written as $\log\, A+\log\, B=0$ or $\log\, AB=1$. This means $AB=1$.

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  • $\begingroup$ Thanks! I got it! $\endgroup$ – Max0815 Feb 14 '19 at 23:24
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It seems that this is a dead end, as I see no other solution other than A=B.

Except, as you now realize, the one and only other option is to have $\log A = - \log B$ (which is possible if $A < 1 < B$ or $B < 1 < A$).

From which it follows $A = e^{\log A} = e^{-\log B} = \frac 1B$ and ... you are back on the right track and you reach a "live" end:

$AB = 1$.

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