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I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:

$$c^{\log(x)}$$ and $$x^{\log(c)}$$

grow at the same rate. When figuring this out I ended up with the expression:

$$\frac{\ln(x)}{\log(x)}=\frac{\ln(c)}{\log(c)}\approx 2.3025$$

This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $\log_{10}(x)$ and $\ln(x)$ ends up being a constant value?

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    $\begingroup$ Indeed the expressions you gave are equal, not just of equal growth: $$c^{\log(x)}=e^{\log(c)\log(x)}=x^{\log(c )}$$ $\endgroup$ – adfriedman Jan 27 '19 at 7:00
  • $\begingroup$ This is just the change of base formula for logarithms that you learn in basic algebra. $\endgroup$ – Brady Gilg Jan 29 '19 at 23:33
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For all $x\neq1$, $x>0$ we have: $$\frac{\ln{x}}{\log{x}}=\frac{\ln{x}}{\frac{\log_ex}{\log_e{10}}}=\log_e10=\ln10.$$

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Some of the answers already provided get close to a full explanation, but not quite.

Recall that if $b > 1$ and $x > 0$, and $$\log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{\rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.

So, why is it that for two bases $a$, $b$, $$\frac{\log_a x}{\log_b x}$$ is a constant not dependent on $x$? The reason is that $\log_a x$ is an exponent, say $y$, such that $a^y = x$; and $\log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$\frac{y}{w} = \log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$\frac{\log_a x}{\log_a b} = \log_b x.$$

Note that only the definition of $\log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.

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Hint: $\ln(10)\approx 2.3025$. Given that information, your conjecture is that $\ln(x)=\ln(10)\log_{10}(x)$. Can you see a way to prove that?

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  • $\begingroup$ I love the effort of guiding people to the answer. $\endgroup$ – Git Gud Jan 27 '19 at 14:25
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If $\ln x = a_x$ and $\log_{10} x = b_x$ then

$e^{a_x} = x$ and $10^{b_x} = x$.

Bear in mind $10 = e^{\ln 10}$ so $10^k = (e^{\ln 10})^k = e^{k\ln 10}$.

So if $10^{k_x} =e^{k_x\ln 10} = x$ then......

By definition $\log_{10} x = k_x$ and $\ln x = k_x\ln 10$ and so.......

$\frac {\ln x}{\log_{10} x} = \frac {k_x\ln 10}{k_x} = \ln 10$.

It's just a conversion constant and shouldn't surprise us.

This is the very basis of the rule $\log_b x = \frac {\log_a x}{\log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).

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Take the logarithm of both expressions: $$ \log(c^{\log x})=\log x\log c \qquad \log(x^{\log c})=\log c\log x $$ So, not only $c^{\log x}$ and $x^{\log c}$ grow at the same rate: they're equal, whatever base of logarithms you use.

For the second part, note that $x=e^{\ln x}=b^{\log_bx}$ by definition. Then $$ \ln x=\log_bx\ln b $$ Therefore, for $x\ne1$, $$ \frac{\ln x}{\log_bx}=\ln b $$

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$log_a b$ = $log_c b \over log_c a$ is a general rule. Thus $log_c a = {log_c b \over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.

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  • $\begingroup$ Ummm.... by your correct rule we have $$\ln(10)=\frac{\log_x(10)}{\log_x(e)}$$ which isn't what we're trying to prove. $\endgroup$ – Rhys Hughes Jan 27 '19 at 6:54
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    $\begingroup$ Thanks. I edited to correct $\endgroup$ – J. W. Tanner Jan 27 '19 at 7:10
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Because $e^{cx}=10^{x}$ for all $x$ and $c:=\ln(10)\approx 2.3025$

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