Why is $\log_a(b)=\frac{\ln(b)}{\ln(a)}$ I know that formula, but I don't understand it.
$\log_a(b)=\frac{\ln(b)}{\ln(a)}$
Thanks.
 A: $$a^{\frac{\ln b}{\ln a}}=e^{\frac{\ln b}{\ln a}\times \ln a}$$
$$=e^{\ln b}$$
$$=b$$
Hence
$$\frac{\ln b}{\ln a}=\log_a{b}$$
A: Remember the definition of $\log_a$: $a^{\log_a(b)}=b$.
We will use the following trick to rewrite $a$: $a=e^{\ln(a)}$, so we can rewrite that first equation: $b=a^{\log_a(b)}=\left(e^{\ln(a)}\right)^{\log_a(b)}=e^{\ln(a)\log_a(b)}$, but we also have $b=e^{\ln(b)}$. So, we have $e^{\ln(a)\log_a(b)}=e^{\ln(b)}$ and by taking $\ln$ of both sides, we get
$$\ln(a)\log_a(b)=\ln(b),$$ 
which can be rewritten to what you asked for.
Note: we didn't have to work with $e$ and $\ln$. Instead, any other number $c>1$ would also work, so the general formula becomes:
$$\log_a(b)=\frac{\log_c(b)}{\log_c(a)}$$ 
A: By definition:
$$
x=\log_a b \quad \iff \quad a^x=b
$$
and
$$
a=e^{\ln a} \qquad b=e^{\ln b}
$$
so:
$$
a^x=b \quad \iff \quad \left(e^{\ln a}\right)^x=e^{x\ln a}=e^{\ln b}
$$
and this gives
$$
x\ln a= \ln b
$$
A: Let $y =  log_a(b)$, then $a^y = b$ .
Take $ln$ of both sides of $a^y = b$.
$ln(a^y) = ln(b)$;
$y$ $ln(a) = ln(b)$;
$y = ln(b)/ln(a)$; 
$y = log_a(b) = ln(b)/ln(a)$.
A: Let $A=\log_xy$, $B=\log_xz$, and $C=\log_zy$. Then $x^A=y$, $x^B=z$ and $z^C=y$. Therefore $x^A=z^C=(x^B)^C=x^{BC}$.
Thus $\log_xy=\log_xz.\log_zy$.
By taking $x=e$, $y=b$, and $z=c$, we have that $\log_eb=\log_ea.\log_ab$. That is $\log_ab=\dfrac{\ln b}{\ln a}$.
