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?
 A: 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 "logarithm base $a$", $\log_a(x)$.
The "guiding formula" 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:
$$\begin{align*}
\log(xy) &= \log(x)+\log(y) &\qquad \ln(xy) &=\ln(x) + \ln(y)\\
\log\left(\frac{x}{y}\right) &= \log(x) - \log(y) &\ln\left(\frac{x}{y}\right) &= \ln(x) - \ln(y)\\
\log(x^a) &= a\log(x) & \ln(x^a) &= a\ln(x)\\
\log(10^x) &= x & \ln(e^x) &= x\\
10^{\log(x)} &= x & e^{\ln(x)} &= x
\end{align*}$$
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).
A: The use of the "ln" abbreviation for natural logarithm is a bad thing because it makes people think that "log" is one thing and "ln" is another thing, and ask what's the difference between the two.
The base-$10$ logarithmic function is a logarithmic function.
The base-$2$ logarithmic function is a logarithmic function.
The base-$e$ logarithmic function is a logarithmic function.
The difference is which number is the base.
Mathematicians writing "$\log x$" usually mean $\log_e x$, also called $\ln x$.
Calculators use $\log x$ to mean $\log_{10} x$.  This is also used in some of the sciences when doing numerical things.
The reason for the importance of base-$10$ logarithms was made obsolete by calculators.  In the early '70s, calculators became widespread.  Before then, many books had tables of base-$10$ logarithms in an appendix.  Suppose you wanted the logarithm of $123$  The table gave you logarithms of numbers between $1$ and $10$, so you found $\log_{10}1.23= 0.089905\ldots$ and concluded that $\log_{10} 123 = 2.089905\ldots\;{}$.  You added $2$ to move the decimal point over 2 places.  That's why base 10 was used: to make that possible.  If you wanted the square root of $7$, you found the logarithm of $7$, divided by $2$, then found the antilogarithm in the same table.  If you wanted to divide $319450231$ by $2673019201$, you found logarithms of both in the table, subtracted, and then found the antilogarithm.  And so on.
The important theoretical question to ask about "$\ln$" is why $e=2.71828182846\ldots$ is the "natural" base to use.  (Has someone posted that question here?)  (When I raise that question and try to answer it in a calculus class, some students ask "Do we HAVE to know this??  Will it be on the test?".  Next time someone does that, I'm going to say "Who cares?".)
A: According to the international standard ISO 31-11 "ln" stands for base-e natural logarithm; "lg" is for base-10 common logarithm; and "lb" is for the base-2 binary one. "log" is a generic notation for a logarithm of an arbitrary base that needs to be specified.
Old Soviet/Russian math/physics textbooks stick to this rule more or less consistently, while English-language physics literature often uses "log" to denote natural logarithm. In defense of the latter I can say that in most cases in physics a logarithm is considered to be an almost constant function. For a theoretical physicist "log(x)" just means "something that doesn't depend too dramatically on x". The exact numeric value of this something is seldom asked for, hence the base is not interesting, hence "log" is just as good as "ln" or "lg" (don't forget these are the same people for whom 2=$\pi$=1).
A: Logs with different bases cross the line y = 0 at x=1 with different slopes (of the tangent to the curve). Natural base e makes this slope equal to 1.
The number is called e after Leonhard Euler, a mathematician that first gave this number a meaning and found its value. Euler worked on a formula for compounding interest. If r is the annualized interest rate and n is a number of compounding intervals per year, the formula for the amount of investment of \$1 after n intervals is: $$(1+\frac r n )^n$$ Euler showed that the limit of this value for infinitely large n is $e^r$ where e is $$\lim_{n\to\infty} (1+\frac 1 n)^n$$ when $n \rightarrow \infty$. It is approximately 2.718.
A: The symbol $\log$, by itself and without other conventions, is meaningless, as is the word "logarithm."
The phrase "logarithm of $x$ to the base $b$" 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 "natural logarithm" is the logarithm to the base $e$, where $e$ is Euler's constant:
$$
\ln(x) = \log_{e}(x).
$$
