1. ## change of base

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

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

$\displaystyle b^y = x$

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

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

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

3. Hello,

Another way :

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