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Given that $\log_ba=\log_ab$, and that $a,b \ne 1$ and $a \ne b$, find $b$ in terms of $a$.

I tried to solve this problem using changing the base of the first part of the equation: $$ \frac{\log_bb}{\log_ba} = \frac{1}{\log_ba} $$ Then multiply both sides by $\log_ba$ and got the following: $$ 1=(\log_ab)^2 $$ Therefore $1=\log_ab$. But this means that $a=b$.

How can I find $b$ in terms of $a$ where $a$ is not equal to $b$?

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  • $\begingroup$ Your title is missing some things. MathJax hint: if you put a backslash before common functions you get the proper font and spacing, so \log x gives $\log x$ compared to log x giving $log x$ $\endgroup$ Commented Sep 20, 2018 at 16:46

3 Answers 3

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$$\log_a b = \dfrac{\ln b}{\ln a}$$

So, we can rewrite the original problem as:

$$\dfrac{\ln a}{\ln b} = \dfrac{\ln b}{\ln a}$$

Then, cross multiply to get:

$$\left(\ln a \right)^2 = \left(\ln b \right)^2$$

Taking square roots, we have:

$$\ln b = \pm \ln a$$

Raise both sides as the power of $e$ and get:

$$b = a^{\pm 1}$$

But, we can eliminate one of those answers (because we know $b \neq a$).

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You're doing very well, until finding that $\log_ab=1$, forgetting that it could also be $$ \log_ab=-1 $$ which implies $b=a^{-1}$.

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Let $a,b >0$.

$a^x=b$; and $b^x= a$.

$(a/b)^x =(b/a) \rightarrow x=-1$.

Then $a^{-1}= b$.

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