The difference between log and ln $$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$

Our professor let's us solve this but i do not understand how $\ln$ works. He says it has same properties with $\log$ but i still don't get it. What's the difference of the two?
The common logarithm is the logarithm base 10. It is the inverse of the exponential function $10^x$. In Calculus and Precalculus classes, it is usually denoted $\log$.
The natural logarithm is the logarithm base $e$. It is the inverse of the exponential function $e^x$. In Calculus and Precalculus classes, it is often denoted $\ln$.
In general, if $a\gt 0$, $a\neq 1$, then the inverse of the function $a^x$ is the &quot;logarithm base $a$&quot;, $\log_a(x)$.
The &quot;guiding formula&quot; is
$$\log_a(b) = r\text{ if and only if }a^r = b.$$
From these, the properties of the logarithmic functions follow:

$\log_a(xy) = \log_a(x)+\log_a(y)$: logarithm of a product is the sum of the logarithms.

Why? Say $\log_a(x) = r$ and $\log_a(y)=s$. That means that $a^r = x$ and $a^s=y$. Then $xy = a^ra^s = a^{r+s}$, so $\log_a(xy) = r+s = \log_a(x) + \log_a(y)$.

$\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)$.

Why? Again, say $\log_a(x) = r$ and $\log_a(y) = s$. Then $a^r = x$, $a^s = y$, so
$\frac{x}{y} = \frac{a^r}{a^s} = a^{r-s}$, which means $\log_a\frac{x}{y}=r-s = \log_a(x)-\log_a(y)$.

$\log_a(x^t) = t\log_a(x)$.

Why? If $\log_a(x)=r$, so that $a^r = x$, then $x^t = (a^r)^t = a^{rt}$, so $\log_a(x^t) = rt = t\log_a(x)$.

$\log_a(a^r) = r$ and $a^{\log_a(x)} = x$. Because $\log_a(x)$ and $a^x$ are inverses of each other.

In particular, $\ln$, which is $\log_{e}$; and using $\log$ for $\log_{10}$, we have these properties:
$$\begin{align*}
\log(xy) &amp;= \log(x)+\log(y) &amp;\qquad \ln(xy) &amp;=\ln(x) + \ln(y)\\
\log\left(\frac{x}{y}\right) &amp;= \log(x) - \log(y) &amp;\ln\left(\frac{x}{y}\right) &amp;= \ln(x) - \ln(y)\\
\log(x^a) &amp;= a\log(x) &amp; \ln(x^a) &amp;= a\ln(x)\\
\log(10^x) &amp;= x &amp; \ln(e^x) &amp;= x\\
10^{\log(x)} &amp;= x &amp; e^{\ln(x)} &amp;= x
\end{align*}$$
It also gives you a way to go back and forth between any logarithm and any other logarithm: if $a$ and $b$ are two bases, both positive, both different from one, what is the relation between $\log_a(x)$ and $\log_b(x)$?
If $\log_b(x)=r$, then $b^r = x$. So
$$\log_a(x)= \log_a(b^r) = r\log_a(b) = \log_b(x)\log_a(b).$$
So we get that
$$\log_b(x) = \frac{\log_a(x)}{\log_a(b)}.$$
As Henning points out below, while $\ln$ is not ambiguous (it always denotes logarithm base $e$), $\log$ is ambiguous and its exact meaning depends on context. In more advanced mathematics courses, it is usual to use it to mean the natural logarithm; in computer science, it is very often used to denote logarithm base $2$. For some applications, it does not matter (for example, when analyzing complexity, since two different logarithms are just scalar multiples of each other).
The symbol $\log$, by itself and without other conventions, is meaningless, as is the word &quot;logarithm.&quot;
The phrase &quot;logarithm of $x$ to the base $b$&quot; has meaning if $b$ and $x$ are positive.  The meaning is defined as follows: the logarithm of $x$ to the base $b$ (denoted by  $\log_{b} (x)$) is the exponent to which $b$ needs to be raised to obtain $x$.
That is, if $y = \log_{b} (x)$, then $x = b^y$.
By convention, the &quot;natural logarithm&quot; is the logarithm to the base $e$, where $e$ is Euler's constant:
$$
\ln(x) = \log_{e}(x).
$$