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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 $\frac{\log(x)}{\log(y)}$?

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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: $a>0$, $b>0$, $a\ne1$, $b\ne1$.

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Hint

$$\log(x)=\log_{10}(x)=\frac{\ln x}{\ln 10}$$

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$\frac{\log(x)}{\log(y)}=\log_y(x)$

$\frac{\ln(x)}{\ln(y)}=\log_y(x)$

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Because $\log x = \frac{\ln x}{\ln(10)}$, so $$\frac{\log x}{\log y} = \frac{\frac{\ln x}{\ln(10)}}{\frac{\ln y}{\ln(10)}} = \frac{\ln x \ln(10)}{\ln y \ln(10)} = \frac{\ln x}{\ln y}.$$

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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 $b$.

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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:$$\frac{\log x}{\log y}=\log_{y}x=\frac{\ln x}{\ln y}$$

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