Why use $\ln x$ and not $\log x$ in most standard intergration formulaes

Question

I wish to understand why we use $\ln x$so much more excessively compared to $\log x$.(log to the base 10)

Taking the following example, I wish to understand why is it so;

$$ ∫dx/(x^2-a^2 )= \frac{1}{2a} \ln⁡|((x-a)/(x+a))| + c \neq \frac{1}{2a} \log_|(x-a)/(x+a)| + c $$

( logx implies logarithm to the base 10 )

Thank you very much.

Answer 1

Due to the standard limit $$ \lim_{x\to 0}\frac{\ln(1+x)}{x}=1 $$ we know that $$ \frac{d}{dx}\ln(x)=\frac{1}{x} $$ with the beautiful unity in the numerator. All other logarithms are proportional to $\ln$ $$ \log_a(x)=\frac{\ln(x)}{\ln(a)}=\text{const}\cdot\ln(x) $$ and satisfy $$ \frac{d}{dx}\log_a(x)=\color{red}{\frac{1}{\ln(a)}}\cdot\frac{1}{x}. $$ It is less appealing in calculations to take care for this extra constant. It makes the logarithm with the base $e$ natural in most calculations in calculus. Moreover, the final conversion from the natural to your particular logarithm of interest is cheap.

Answer 2

Well:

$$\log_e\left(b\right):=\ln\left(b\right)$$

With $e$ is the base of natural logarithm and $e=2.718281828459045...$

Answer 3

Convenience, basically; consider:

• $\frac{d}{dx}\log_{10}x=\frac{1}{\ln 10}\frac{1}{x}$
• $\frac{d}{dx}\ln x=\frac{1}{x}$

Effectively, the base-$e$ logarithm leads to neater, more natural calculations in the framework of differentiation and integration, and is thus favoured in this context. It's similar to how [if you've seen it] we use radians rather than degrees, when using trigonometric functions in calculus - calculations become far more natural through this choice.

Answer 4

In calculus, we use the natural logarithm because it's natural!

By this, I mean that any other choice of base would require to introduce a scaling constant in our formula. We can do this, but as a convention mathematicians don't (as it's more work for very little benefit).

An easy example of this isn't with logarithms, but with exponential (which are the inverse of the logarithim), where:

$$\frac{d} {dx} a^x = \ln(a) a^x$$ This is just how the calculation works out. So, when we choose a base for the exponential, it makes sense to let $a = e$, as $\ln(e) =1$ so the scaling factor disappears.

This is the heart of the matter. If we choose a non $e$ base, we need to keep track of coefficients more.

Answer 5

It helps to simplify calculations some. The derivative of $log_a(x)$ is $\frac{1}{xln(a)}$. When the base is $a = e$, the natural log term becomes $1$, so we get that the derivative of $ln(x)$ is $\frac{1}{x}$. In the example you gave (as well as many others), you convert the integrand into something of the form $\frac{du}{u}$ using some form of u-substitution (and possibly partial fractions), so that the antiderivative is the natural logarithm of $u$. You could always choose to use $log_a(x)$ as well, but it carries a constant with it. It's simply standard notation (as well as vaguely nice for calculations) to use natural logarithm instead.