# Thread: Proving the log equation

1. ## Proving the log equation

logb(a)+logc(b)+loga(c)=1/loga(b) + logb(c) + 1/logc(a)

Hrmm.. im thinking loga(a)/loga(b), how does that equal to logb(a) ?

Thanks

2. Hello smmmc
Originally Posted by smmmc
logb(a)+logc(b)+loga(c)=1/loga(b) + logb(c) + 1/logc(a)

Hrmm.. im thinking loga(a)/loga(b), how does that equal to logb(a) ?

Thanks
The definition of a logarithm is as follows:
The log of a number, to a certain base, is that power to which the base must be raised to get the number.
So, by definition, if
$x=\log_a(b)$
then
$a^x=b$
Therefore, using the laws of indices:
$a=b^{\frac1x}$
and so, using the definition of a log once again:
$\log_b(a)=\frac1x$
$=\frac{1}{\log_a(b)}$

3. i understand up to a=b^(1/x)

but still dont get why it equals 1/loga(b)

so from logb(a)=1/x, i just see this as b^(1/2)=a

?
Thanks

4. Dear smmmc,

Think it this way,

$log_{a}b=x$

Grandad has shown, $log_{b}a=\frac{1}{x}$

Therefore, by substitution $log_{b}a=\frac{1}{log_{a}b}$

Hope this helps.

5. Hello smmmc

OK. I see that you get it to here:
Originally Posted by smmmc
i understand up to a=b^(1/x)
But I'm not sure what you mean by this:
...so from logb(a)=1/x, i just see this as b^(1/2)=a
I understand the difficulty you're having with logs, because the definition of a logarithm is a bit 'inside-out' somehow. But it's important for you to get your head around it, so here's a bit more explanation.

If $a=b^{\frac1x}$, then $\frac1x$ is the power to which $b$ must be raised in order to give the answer $a$. Right?

So, looking back to the definition of a log that I gave you at the beginning, this means that $\frac1x = \log_b(a)$.

Well if $\log_b(a) =\frac1x$ and $x = \log_a(b)$, then, by substitution, $\log_b(a) = \frac{1}{\log_a(b)}$.

6. Originally Posted by smmmc
i understand up to a=b^(1/x)

but still dont get why it equals 1/loga(b)

so from logb(a)=1/x, i just see this as b^(1/2)=a
1/x, not 1/2. I assume that was a typo.
?
Thanks

### logb a * logc b * loga c

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