Why is the Ratio of $\ln(x)$ and $\log(x)$ a constant?

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

For all $x\neq1$, $x>0$ we have: $$\frac{\ln{x}}{\log{x}}=\frac{\ln{x}}{\frac{\log_ex}{\log_e{10}}}=\log_e10=\ln10.$$

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?