Suppose $A$ and $B$ are positive real numbers for which $\log_{A} B = \log_{B} A$ . If neither $A$ nor $B$ is $1$ and $A$ $\not\equiv B$, find the value of $AB$.
We can rewrite as
$A^{\log_{A} B} = B$.
$B^{\log_{B} A} = A$.
$(B^{\log_{B} A})^{A^{\log_{A} B}} =B$.
$B^{2 (\log_{B} A)} =B$.
$B^{2 (\frac{1}{2})} =B$.
I'm unable to deduce a correct solution with $\frac{1}{2}$. I wanted to try a different approach by converting the powers to multiplication but can't figure out the base. i.e.
${\log_{A} B (\log_{?}A)} $.
$\endgroup$3 Answers
$\begingroup$To change from the base $b$ to the base $c$ in the logarithm below$$\log_{b}a$$we can use the following formula$$\log_{b}a=\frac{\log_{c}a}{\log_{c}b}$$
So, to change from the base $B$ to base $A$:$$\log_{B}A=\frac{\log_{A}A}{\log_{B}A}$$and, since $\log_{A}A=1$, we get$$\log_{B}A=\frac{1}{\log_{A}B}$$and$$\log_{B}A=\log_{A}B$$then$$(\log_{A}B)^{2}=1\Rightarrow\log_{A}B=\pm1$$Therefore,$$A=B$$or$$A=B^{-1}$$
$\endgroup$ 4 $\begingroup$From $A^{\log_A B} = B$ and $B^{\log_B A} = A$ we can write:
$$(A^{\log_A B})^{\log_B A} = A = A^1$$
Hence $(\log_A B) (\log_B A) = 1$.
But we are given $\log_A B = \log_B A$. Hence $(\log_A B)^2 = 1$.
If $\log_A B = 1$ then $A = B$, which contradicts an assumption.
Hence $\log_A B = -1$. What does that imply for $AB$?
$\endgroup$ $\begingroup$\begin{eqnarray*} \log_A(B) = \log_B(A) \\ \frac{\ln(B)}{\ln(A)}=\frac{\ln(A)}{\ln(B)} \\ (\ln(B))^2=(\ln(A))^2 \\ \ln(B)=-\ln(A) \text{ or ...} \\ \ln(AB)=0. \end{eqnarray*}S0 $AB=?$
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