# change of base

• May 13th 2009, 02:42 PM
VonNemo19
change of base
prove that $\log{b}x=\frac{\log{a}x}{\log{a}b}$. I'm having a little trouble getting started.
• May 13th 2009, 02:52 PM
skeeter
Quote:

Originally Posted by VonNemo19
prove that $\log{b}x=\frac{\log{a}x}{\log{a}b}$. I'm having a little trouble getting started.

$y = \log_b{x}$

$b^y = x$

$\log_a{b^y} = \log_a{x}$

$y\log_a{b} = \log_a{x}$

$y = \frac{\log_a{x}}{\log_a{b}}$
• May 14th 2009, 10:31 AM
Moo
Hello,

Another way :

$\log_b x=\frac{\ln(x)}{\ln(b)}=\frac{\ln(x)}{{\color{red} \ln(a)}} \cdot \frac{{\color{red}\ln(a)}}{\ln(b)}=\log_ax\cdot\fr ac1{\log_ab}$