Xrhrft

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?

For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$

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 $10^k = (e^{ln 10})^k = e^{kln 10}$.

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

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

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

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).

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?

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

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 $(b^m)^n = b^{mn}$.