Why $\frac{\log(x)}{\log(y)}$ gives the same value as $\frac{\ln(x)}{\ln(y)}$ If $x=16384$ and $y=2$
$\ln(x)=9.704$
$\ln(y)=0.6931$
$\log(x)=4.2144$
$\log(y)=0.3010$
If we divide $\frac{\ln(x)}{\ln(y)}$ we get $14$ and same answer for $\frac{\log(x)}{\log(y)}$.
So can anyone tell me the concept behind this? Why does dividing $\frac{\ln(x)}{\ln(y)}$ give the same result as $\frac{\log(x)}{\log(y)}$?
 A: The logarithm in base $a$ satisfies
$$
\log_a r = s
\qquad
\text{if and only if}
\qquad
a^s=r
$$
Now, compute the logarithm in base $b$ of $a^s=r$:
$$
\log_b a^s=\log_b r
$$
that is
$$
s\log_b a=\log_b r
$$
so
$$
\log_b r= \log_a r \cdot \log_b a.
$$
Now
$$
\frac{\log_b x}{\log_b y}
=
\frac{\log_a x \cdot \log_b a}{\log_a y \cdot \log_b a}
=
\frac{\log_a x}{\log_a y}
$$
for any $a,b$ such that the logarithm makes sense: $a>0$, $b>0$, $a\ne1$, $b\ne1$.
A: Hint
$$\log(x)=\log_{10}(x)=\frac{\ln x}{\ln 10}$$
A: $\frac{\log(x)}{\log(y)}=\log_y(x)$
$\frac{\ln(x)}{\ln(y)}=\log_y(x)$
A: Because $\log x = \frac{\ln x}{\ln(10)}$, so 
$$\frac{\log x}{\log y} = \frac{\frac{\ln x}{\ln(10)}}{\frac{\ln y}{\ln(10)}} = \frac{\ln x \ln(10)}{\ln y \ln(10)} = \frac{\ln x}{\ln y}.$$
A: Let $x=y^k$. Then $\log_b(x)=k\cdot\log_b(y)$, so
$$
\frac{\log_b(x)}{\log_b(y)} = k
$$
which is constant relative to the base $b$.
A: For each $h>0$ with $h\neq 1$:
$$y^{\log_{y}x}=x=h^{\log_{h}x}=\left(h^{\log_{h}y}\right)^{\frac{\log_{h}x}{\log_{h}y}}=y^{\frac{\log_{h}x}{\log_{h}y}}$$
hence: $$\log_{y}x=\frac{\log_{h}x}{\log_{h}y}$$ 
Taking $h=10$ and $h=e$ you find:$$\frac{\log x}{\log y}=\log_{y}x=\frac{\ln x}{\ln y}$$
