# Thread: How do I prove this Log?

1. ## How do I prove this Log?

For a>1, b>1, show that (logab)(logba) = 1.

The variable after the log is the base.

How do I go about showing this?

Logbx = Logax / Logab

(again, the variable after the log is the base, i don't know how to write it properly on this forum)

and, Logtb = 1 / Logbt

(variable after log is base)

How do I approach / solve this problem?

2. $\displaystyle\log_bx=\frac{\log_ax}{\log_ab}$.
For $x=a$ the equality becomes
$\log_ba=\frac{\log_aa}{\log_ab}=\frac{1}{\log_ab}$

3. Okay...but how do I show that (logab)(logba) = 1?

4. So, $\log_ba=\frac{1}{\log_ab}$
Multiply the equality by $\log_ab$.
You get $\log_ab\cdot\log_ba=1$

5. $\log_b a = x$ and $\log_a b = y$.
We want to show $xy=1$.
But this means (by definition),
$a=b^x \mbox{ and }a^y = b$.
Thus,
$a=b^x\implies (a^y)=(b^x)^y\implies a^y = b^{xy}\implies b =b^{xy}\implies xy=1$

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# show that logaB logba=1

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