# 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?

• Indeed the expressions you gave are equal, not just of equal growth: $$c^{\log(x)}=e^{\log(c)\log(x)}=x^{\log(c )}$$ Jan 27, 2019 at 7:00
• This is just the change of base formula for logarithms that you learn in basic algebra. Jan 29, 2019 at 23:33

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?

• I love the effort of guiding people to the answer. Jan 27, 2019 at 14:25

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).

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 $$x\ne1$$, $$\frac{\ln x}{\log_bx}=\ln b$$

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

• 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. Jan 27, 2019 at 6:54
• Thanks. I edited to correct Jan 27, 2019 at 7:10

Because $$e^{cx}=10^{x}$$ for all $$x$$ and $$c:=\ln(10)\approx 2.3025$$