change of base

**VonNemo19** · May 13th 2009, 01:42 PM

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

**skeeter** · May 13th 2009, 01:52 PM

Originally Posted by VonNemo19

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

$y = \log_b{x}$

$b^y = x$

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

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

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

**Moo** · May 14th 2009, 09:31 AM

Hello,

Another way :

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

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