# growth function and prove

• Nov 26th 2009, 11:46 AM
isiksoy7
growth function and prove
[IMG]file:///C:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]Can you help for the functions and prove at the attachment ?
• Nov 26th 2009, 03:07 PM
tonio
Quote:

Originally Posted by isiksoy7
[IMG]file:///C:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]Can you help for the functions and prove at the attachment ?

This is basic logarithms stuff, I think: use that $\displaystyle \log_ax=\frac{\ln x}{\ln a}$ , so

$\displaystyle \frac{f(n)}{g(n)}=\frac{\frac{\ln n}{\ln a}}{\frac{\ln n}{\ln b}}=\frac{\ln b}{\ln a}$

Tonio
• Nov 26th 2009, 03:17 PM
emakarov
The key is to show that $\displaystyle \log_an=C_1\log_bn$ and $\displaystyle \log_bn=C_2\log_an$ for some constants $\displaystyle C1, C2$.

Now logrithm is not such a scary beast. Personally, I remember only three properties of logarithms.

(1) Definition: $\displaystyle \log_an=x$ iff $\displaystyle a^x=n$
(2) $\displaystyle \log_a(x^y)=y\log_ax$
(3) $\displaystyle \log_a(xy)=\log_ax+\log_ay$.

It's not difficult to deduce (2) and (3) from (1), so the main thing you need to remember is that logarithm is an inverse function to power: $\displaystyle a^{\log_ax}=x$ and $\displaystyle \log_a(a^x)=x$.

So, how to express $\displaystyle \log_an$ through $\displaystyle \log_bn$? Let $\displaystyle \log_an=x$. By (1), $\displaystyle a^x=n$. We need to use $\displaystyle \log_bn$, so let's take $\displaystyle \log_b$ of both sides: $\displaystyle \log_b(a^x)=\log_bn$. By (2), $\displaystyle x\log_ba=\log_bn$. Recalling the definition of $\displaystyle x$, we get $\displaystyle \log_ba\log_an=\log_bn$.

It's also easy to remember this last formula. Here is a mnemonic (not mathematical) explanation: you go from $\displaystyle b$ to $\displaystyle a$ ($\displaystyle \log_ba$), then from $\displaystyle a$ to $\displaystyle n$ ($\displaystyle \log_an$) and the result is from $\displaystyle b$ to $\displaystyle n$ ($\displaystyle \log_bn$).
• Nov 27th 2009, 02:16 AM
isiksoy7