# For distinct positive reals $A$ and $B$, neither equal to $1$, such that $\log_A B = \log_B A$, find $AB$.

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

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 $$B=A^{-1}$$.

• Oh, a generalization of it, thanks! – Max0815 Feb 14 '19 at 23:38

$$x^{2}=y^{2}$$ 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 $$AB=1$$.

• Thanks! I got it! – Max0815 Feb 14 '19 at 23:24

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:

$$AB = 1$$.