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$$\dfrac{1}{2}\ln(x+7)-(2 \ln x+3 \ln y)$$

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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    $\begingroup$ 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. $\endgroup$ – The Chaz 2.0 Dec 11 '11 at 22:20
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    $\begingroup$ 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? $\endgroup$ – Arturo Magidin Dec 11 '11 at 22:23
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    $\begingroup$ The difference is that $\ln$ is for children, and $\log$ is for grownups. $\endgroup$ – Gerry Myerson Dec 11 '11 at 22:46
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    $\begingroup$ @GerryMyerson : You be nice or I'll whack you across the knuckles with my slide rule. $\endgroup$ – steven gregory Nov 3 '15 at 18:39
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    $\begingroup$ honestly why isn't it nl x ? $\endgroup$ – futurebird Mar 21 '18 at 23:16
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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).

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    $\begingroup$ 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 $\log_2$. $\endgroup$ – hmakholm left over Monica Dec 11 '11 at 22:49
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    $\begingroup$ Good point; I guess I've been teaching too many Calculii lately... $\endgroup$ – Arturo Magidin Dec 11 '11 at 22:59
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    $\begingroup$ Don't even assume that log = base2 for software; TensorFlow, for example, has a log() function that "Computes natural logarithm of x element-wise." The use of "log" appears to be intentionally annoying across fields. $\endgroup$ – James Moore Nov 4 '16 at 15:38
  • $\begingroup$ Can I say that, logarithmic function does not have its own definition except that it is originated as inverse of exponential(e^x) function? $\endgroup$ – overexchange Nov 24 '16 at 13:34
  • $\begingroup$ Expanding on Henning's comment, even "ln" is ambiguous in the wider community because it's interpreted as short for "logarithm" versus the esoteric "logarithmus naturalis" (in which case "nl" or "nlog" would be better). And because there are several common yet distinct bases used when doing logarithm work, which people tend to omit to specify. As a programmer, when writing code I like to clearly distinguish using function aliases like log2(), log10(), loge(). Or a single overloaded function log(), which when given two parameters takes the first as the base, like log(2,x). $\endgroup$ – Beejor Apr 8 '19 at 21:13
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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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    $\begingroup$ I had a professor who always answered "Will it be on the test?" with "Now it will definitely be"... $\endgroup$ – Arturo Magidin Dec 11 '11 at 23:25
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    $\begingroup$ "The important theoretical question to ask about "ln" is why e=2.71828182846… is the "natural" base to use." That question is discussed here. $\endgroup$ – Austin Mohr Dec 12 '11 at 0:21
  • $\begingroup$ Since the default base of log can vary between and even within fields, seems a good rule of thumb is to treat ln as loge (of course), and log as unknown (re: base-2/10/e/whatever) until you confirm the context. If calculating or programming, check a test result before making assumptions. And in the latter case, it may be best to simply ignore a language's built-in log function and use your own so that your code is more portable. $\endgroup$ – Beejor Apr 8 '19 at 21:27
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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.

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