<s>[INST] Why is ln(x) bigger than log(x)? I couldn't find any results. I would like to know why $\ln x$ stays higher than $\log x$. Does it have to do with ase $10$ of $\log$? base $e$ with $\ln x$?
 [/INST] Take two different bases: $2$ and $4$.  Now let's look at what exponent is required to equal some numbers (this is what the logarithm computes):
\begin{align}
2^? = 16 \rightarrow 2^4 = 16 && 4^? = 16 \rightarrow 4^2 = 16\\
2^? = 64 \rightarrow 2^6 = 64 && 4^? = 64 \rightarrow 4^3 = 64 \\
2^? = 1024 \rightarrow 2^{10} = 1024 && 4^? = 1024 \rightarrow 4^5 = 1024
\end{align}
Clearly the smaller base requires a larger exponent to make the same value.  Thus since $e < 10$, if $e^a = 10^b$ then $a > b$ and thus $\ln(x) > \log_{10}(x)$ (for some values of $x$).
Let's look at some different numbers with the same two bases above:
\begin{align}
2^? = \frac{1}{16} \rightarrow 2^{-4} = \frac{1}{16} && 4^? = \frac{1}{16} \rightarrow 4^{-2} = \frac{1}{16}\\
2^? = \frac{1}{64} \rightarrow 2^{-6} = \frac{1}{64} && 4^? = \frac{1}{64} \rightarrow 4^{-3} = \frac{1}{64} \\
2^? = \frac{1}{1024} \rightarrow 2^{-10} = \frac{1}{1024} && 4^? = \frac{1}{1024} \rightarrow 4^{-5} = \frac{1}{1024}
\end{align}
In this case, first, notice the absolute value of the exponent is still greater, but since they are negative, the smaller number requires a "smaller" exponent (a "more negative" exponent).
So when the exponent is negative, meaning $\log_a(x) < 0$ and thus $x < 1$, the smaller base requires a "smaller" exponent (really a more negative exponent).  The general rule of :
$$
\left|\log_a(x)\right| \geq \left|\log_b(x)\right|, a > 1, b > 1, a \leq b
$$
If $a$ and/or $b$ is between $0$ and $1$ then it gets tricky since:
$$
0 < a < 1 \rightarrow \begin{cases}
\log_a(x) < 0 & x > 1 \\
\log_a(x) > 0 & x < 1
\end{cases}
$$
To figure out how the absolute values would compare requires looking at the inverse of the bases.  Let's continue to assume that $\frac{1}{a} < b$ and $a, b > 0$:


*

*If $a < b$ then the same rules would apply:
$$
\left|\log_{\frac{1}{a}}(x)\right| \geq \left|\log_b(x)\right|
$$

*If $a > b$ then the opposite rules would apply:
$$
\left|\log_{\frac{1}{a}}(x)\right| \leq \left|\log_b(x)\right|
$$


If both $a$ and $b$ are between $0$ and $1$ then, since the smaller number would mean the larger inverse (i.e. $\frac{1}{a} < \frac{1}{b} \rightarrow b < a$), the smaller number would make the smaller logarithm.
Here is a summarization, assuming $a, b > 1$ and $a < b$:
\begin{align}
\left|\log_a(x)\right| \geq& \left|\log_b(x)\right| \\
\left|\log_{\frac{1}{a}}(x)\right| \geq& \left|\log_b(x)\right| \\
\left|\log_a(x)\right| \geq& \left|\log_{\frac{1}{b}}(x)\right| \\
\left|\log_{\frac{1}{a}}(x)\right| \geq& \left|\log_{\frac{1}{b}}(x)\right|
\end{align}
</s>