# 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?

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ln denotes the natural log, which has base $e$. Just "log" in this context denotes a log with base $10$. Google "mathematical constant e" for more info. –  The Chaz 2.0 Dec 11 '11 at 22:20
The formula is ambiguous. Do you mean $$\frac{1}{2}\ln(x+7)\text{ or }\frac{1}{2\ln(x+7)}$$in the first summand? –  Arturo Magidin Dec 11 '11 at 22:23
@MaX: How do you know which one was meant? –  Arturo Magidin Dec 11 '11 at 22:38
The difference is that $\ln$ is for children, and $\log$ is for grownups. –  Gerry Myerson Dec 11 '11 at 22:46
@Arturo:It was a guess on my side, but now i see your point, rolled back to the the OP's version. –  Quixotic Dec 12 '11 at 8:54

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?".)

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I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"... –  Arturo Magidin Dec 11 '11 at 23:25
"The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here. –  Austin Mohr Dec 12 '11 at 0:21

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:

1. $\log_a(xy) = \log_a(x)+\log_a(y)$: 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 $\log_a(xy) = r+s = \log_a(x) + \log_a(y)$.

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

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 $\log_a\frac{x}{y}=r-s = \log_a(x)-\log_a(y)$.

3. \log_a(x^t) = t\log_a(x)$$. Why? If \log_a(x)=r, so that a^r = x, then x^t = (a^r)^t = a^{rt}, so \log_a(x^t) = rt = t\log_a(x). 4. \log_a(a^r) = r and a^{\log_a(x)} = x. Because \log_a(x) and a^x are inverses of each other. In particular, \ln, which is \log_{e}; and using \log for \log_{10}, we have these properties:$$\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). - Beware that$\log$does not unambiguously mean the base-10 logarithm, but rather "the logarithm that we usually use". In many areas of higher mathematics,$\log$means the natural logarithm and the$\ln$notation is seldom seen. And computer scientists routinely use$\log$to mean$\log_2\$. –  Henning Makholm Dec 11 '11 at 22:49
Good point; I guess I've been teaching too many Calculii lately... –  Arturo Magidin Dec 11 '11 at 22:59