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
Hello smmmcThe 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
$\displaystyle x=\log_a(b)$then$\displaystyle a^x=b$Therefore, using the laws of indices:$\displaystyle a=b^{\frac1x}$and so, using the definition of a log once again:$\displaystyle \log_b(a)=\frac1x$Grandad$\displaystyle =\frac{1}{\log_a(b)}$
Hello smmmc
OK. I see that you get it to here:But I'm not sure what you mean by this: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....so from logb(a)=1/x, i just see this as b^(1/2)=a
If $\displaystyle a=b^{\frac1x}$, then $\displaystyle \frac1x$ is the power to which $\displaystyle b$ must be raised in order to give the answer $\displaystyle a$. Right?
So, looking back to the definition of a log that I gave you at the beginning, this means that $\displaystyle \frac1x = \log_b(a)$.
Well if $\displaystyle \log_b(a) =\frac1x$ and $\displaystyle x = \log_a(b)$, then, by substitution, $\displaystyle \log_b(a) = \frac{1}{\log_a(b)}$.
Grandad