# Please explain this simple rule of logarithms to me Right I know this one is simple and I know that

Keenan Santos 2022-07-06 Answered
Please explain this simple rule of logarithms to me
Right I know this one is simple and I know that I just need a push to make it sink in in my head..
I am studying control systems and in one of the tutorial examples the tutor says
Show that
$20\mathrm{log}\left(1/x\right)=-20\mathrm{log}\left(x\right)$
I know that when you have a divide or a multiply with logarithms you add them and subtract them but for my own understanding I just need someone to like slowly show me how this works..
If I take the log of the numerator I have $20\mathrm{log}\left(1\right)=0$ but I don't know where to go from here.. So do I now just take the log of the denominator and as the numerator was zero it is just minus whatever the log of the denominator is... Getting myself a bit muddled.. Thanks
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Caiden Barrett
First of all, forget the $20$ in front, all you have to do is see that
$\mathrm{log}\left(1/x\right)=-\mathrm{log}\left(x\right).$
You accept the fact that
$\mathrm{log}\left({a}^{b}\right)=b\cdot \mathrm{log}\left(a\right)$
and use the fact that $1/x={x}^{-1}$, giving you
$\mathrm{log}\left(1/x\right)=\mathrm{log}\left({x}^{-1}\right)=\left(-1\right)\mathrm{log}\left(x\right)=-\mathrm{log}x$
You use the fact that
$\mathrm{log}\left(a/b\right)=\mathrm{log}a-\mathrm{log}b$
and that $\mathrm{log}1=0$, giving you
$\mathrm{log}\left(1/x\right)=log\left(1\right)-\mathrm{log}\left(x\right)=0-\mathrm{log}\left(x\right)=-\mathrm{log}\left(x\right)$
You use the fact that $\mathrm{log}1=0$, that $1=x\ast \left(1/x\right)$ and that
$\mathrm{log}\left(ab\right)=\mathrm{log}a+\mathrm{log}b,$
giving you
$0=\mathrm{log}1=\mathrm{log}\left(x\ast \left(1/x\right)\right)=\mathrm{log}x+log\left(1/x\right).$
From the equation
$0=\mathrm{log}x+\mathrm{log}\left(1/x\right),$
you get
$\mathrm{log}\left(1/x\right)=-\mathrm{log}x$
Note that all these derivations can be transformed into each other and are equivalent. I am presenting them all because everybody looks to logarighms in his own way, so make your pick of the favorite.

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Grimanijd
Essential for logarithms is equality
${g}^{{\mathrm{log}}_{g}a}=a$
This under the conditions $a>0$, $g>0$ and $g\ne 1$
Asking: 'to what power must $3$ ($=g$) be raised to get $81$ ($=a$) as outcome?' is the same thing as asking: 'what is the logarithm of $81$ on base of $3$?' The answer is clearly $4$. We have ${3}^{4}=81$ and equivalent is the expression: ${\mathrm{log}}_{3}81=4$
Note that:
${10}^{20\mathrm{log}\left(\frac{1}{x}\right)}={\left({10}^{\mathrm{log}\left(\frac{1}{x}\right)}\right)}^{20}={\left(\frac{1}{x}\right)}^{20}$
so the fact that ${\left(\frac{1}{x}\right)}^{20}={x}^{-20}$ tells us that $10$ is raised in these cases to the same power.
Our conclusion is:
$20\mathrm{log}\left(\frac{1}{x}\right)=-20\mathrm{log}x$
Every rule concerning logarithms can derived likewise. For instance:
${g}^{{\mathrm{log}}_{g}a+{\mathrm{log}}_{g}b}={g}^{{\mathrm{log}}_{g}a}×{g}^{{\mathrm{log}}_{g}a}=ab={g}^{{\mathrm{log}}_{g}ab}$
resulting in:
${\mathrm{log}}_{g}a+{\mathrm{log}}_{g}b={\mathrm{log}}_{g}ab$

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