Why is the Ratio of $\ln(x)$ and $\log(x)$ a constant? 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?
 A: 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?
A: 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 $(b^m)^n = b^{mn}$.
A: 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).
A: 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 $x\ne1$,
$$
\frac{\ln x}{\log_bx}=\ln b
$$
A: For all $x\neq1$, $x>0$ we have: $$\frac{\ln{x}}{\log{x}}=\frac{\ln{x}}{\frac{\log_ex}{\log_e{10}}}=\log_e10=\ln10.$$
A: $log_a b$ = $log_c b \over log_c a$ is a general rule. 
Thus $log_c a = {log_c b \over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.  
A: Because $e^{cx}=10^{x}$ for all $x$ and $c:=\ln(10)\approx 2.3025$
