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

Question

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

Answer 1

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 $B=A^{-1}$.

Answer 2

$x^{2}=y^{2}$ 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 can be written as $\log\, A+\log\, B=0$ or $\log\, AB=0$. This means $AB=1$.

Answer 3

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:

$AB = 1$.

Answer 4

I found a really simple solution using only the definition of log and basic exponential rules.

Let $\log_{A}B = x = \log_{B}A$. Using the definition of log, $$ B^x = A\,\,\text{and}\,\,A^x = B.\tag{1} $$ $$ A^x = B \implies (A^x)^x = B^x \implies A^{x^2} = B^x. \tag{2} $$ Using equations $(1)$ and $(2)$, $$ A^{x^2} = A $$ Since $A$ is not equal to $1$, and the bases are identical, we can equate the powers: $$ x^2 = 1 \implies x =\pm 1. $$

Case 1: $x = 1$

By equation $(1)$, $A = B$, which is not allowed by the question.

Case 2: $x = -1$

$A$ and $B$ become reciprocals of each other, and $\boxed{AB = 1}$.