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

$$20\log(1/x) = -20\log(x)$$

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 $20\log(1) = 0$ 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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First of all, forget the $20$ in front, all you have to do is see that

$$\log (1/x) = - \log(x).$$ Multiplying that with $20$ will give you what you neeed. For this, I have a few options:

• You accept the fact that $$\log(a^b) = b\cdot \log(a)$$ and use the fact that $1/x=x^{-1}$, giving you $$\log (1/x) = \log(x^{-1}) = (-1)\log(x) = -\log x$$

• You use the fact that $$\log(a/b) = \log a - \log b$$ and that $\log 1 = 0$, giving you $$\log (1/x) = log(1) - \log(x) = 0 - \log(x) = -\log(x)$$

• You use the fact that $\log 1 = 0$, that $1 = x* (1 / x)$ and that $$\log(ab) = \log a + \log b,$$ giving you $$0 = \log 1 = \log(x*(1/x)) = \log x + log(1/x).$$ From the equation $$0 = \log x + \log (1/x),$$ you get $$\log(1/x) = - \log x$$

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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This is an amazing answer I am adding it to my Evernote maths notebook on logarithms, thankyou – lilSebastian Jan 16 '14 at 11:45

Essential for logarithms is equality $$g^{\log_{g}a}=a$$ This under the conditions $a>0$, $g>0$ and $g\neq1$.

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: $\log_{3}81=4$.

Note that: $$10^{20\log\left(\frac{1}{x}\right)}=\left(10^{\log\left(\frac{1}{x}\right)}\right)^{20}=\left(\frac{1}{x}\right)^{20}$$ and: $$10^{-20\log x}=\left(10^{\log x}\right)^{-20}=x^{-20}$$ 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: $$20\log\left(\frac{1}{x}\right)=-20\log x$$

Every rule concerning logarithms can derived likewise. For instance: $$g^{\log_{g}a+\log_{g}b}=g^{\log_{g}a}\times g^{\log_{g}a}=ab=g^{\log_{g}ab}$$ resulting in: $$\log_{g}a+\log_{g}b=\log_{g}ab$$

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You have : $0 = 20\times\log(1) = 20\times\log(x\times\frac1{x})$

But $\log(x\times \frac1{x})=\log(x)+\log(\frac1{x})$.

So $0=20(\log(x)+\log(\frac1{x}))$ and you have your result.

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you know that

$log(x^2)=2*log(x)$

now what about $1/x$? can we say that $1/x=x^{-1}$ ?

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This makes perfect sense. Thank you so much.. – lilSebastian Jan 16 '14 at 10:05
you are welcome,thanks very much – dato datuashvili Jan 16 '14 at 10:06

log(1/x) =log(1)-log(x) because if y=log(x) it implied 10^y =x therfore 20log(1/x) = 20(log(1)-log(x))=-20log(x)

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Consider a number like 1,000,000. Square it, you get 1,000,000,000,000. Notice that when you squared it, the number of zeroes doubled. One rough way of looking at logarithms is "the number of digits" of a number (in base 10 anyway, ymmv).

You can derive formally what we show above informally:

$$\log(x^n) = n \, \log(x)$$

Cube the result, you get triple the number of digits, etc...In this case $n=-1$.

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