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 $\ln x$?
 A: We have the change of base formula
$$\log_b(x)=\frac{\log_a(x)}{\log_a(b)}$$
This gives rise to the equation $\log_{10}(e)\log_e(x)=\log_{10}(x)$ and we note that $\log_{10}(e)<1$. I assume you mean measuring from $x=1$ onwards, because they meet and cross at that $(1,0)$
A: The idea of the function $y=log_{10}(x)$ is to find $y$ so that $10^y=x$.
If you take an example $log_{10}100$ compared to $log_2 100$ 
$log_{10}100=2$, you have to square $10$ to get $100$
$log_{2}100=6.6$, you have to multiply $2$ by itself more than $6$ times to get $100$. It should be clear that since $2$ is smaller than $10$ you have to multiply it by itself more times to get $100$. In the same way $e$ is smaller than $10$ so you'll have to multiply it by itself more times to get any number (any bigger than $1$).
A: 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}
A: An alternative approach: $$\log_{10}(x) = \int_1^x \frac{1}{t \cdot(\ln10)} dt < \int_1^x \frac{1}{t} dt = \ln(x)$$
for $x>1$.  This is clear since the area under the graph of  $\frac{1}{t \cdot(\ln10)}$ is strictly contained within the area of the graph of $\frac{1}{t}$ for $t>1$.
A: TLDR:
$$(\frac{d}{dx}10^x>\frac{d}{dx}e^x)$$
$$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$
$$(\frac{1}{\frac{d}{dx}10^x}<\frac{1}{\frac{d}{dx}e^x})$$

Let me offer a quick and intuitive approach.
As we know from this famous reciprocal relationship: 
$$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$
Which we shall use to help with interpreting the exponential function's natural reciprocal relationships:
$$y=f(x)=e^x$$
$$x=g(y)=log_ey$$
Hence:
$$y=f(x)=(\frac{10}{e}e)^x=(\frac{10}{e})^x(e)^x$$
$$x=g(y)=log_{10}y$$
And in where:
$$\frac{d}{dx}a^xe^x=(\frac{d}{dx}a^x)(e^x)$$
Conclusion:
-$e^x$ and its 1st order derivative is equal to $e^x$.
-Any function proportional to $e^x$ of the form $a^xe^x$ will have a 1st order derivative that is proportional to $e^x$, being $(\frac{d}{dx}a^x)(e^x)$.
-For the case of $(\frac{10}{e}e)^x$, its 1st order derivative is proportional to $e^x$ by a factor ($\frac{d}{dx}a^x$) that is monotically increasing as $\frac{10}{e}=a>1$ for as $10>e$ - $(\frac{d}{dx}10^x>\frac{d}{dx}e^x)$.
-Therefore, taking advantage of $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$, we have $(\frac{1}{\frac{d}{dx}10^x}<\frac{1}{\frac{d}{dx}e^x})$.

Supplement: 
For $x<1$, $log_{10}x>log_ex$
