Natural logarithms

In a nutshell

The natural logarithm, $\ln(x)$ , is the logarithm with base $e$ , and therefore the functions $e^x$ and $\ln(x)$ are inverses to each other.

The graph of the equation $y = \ln(x)$

A consequence of the functions $e^x$ and $\ln(x)$ being inverse to each other is that the graphs of the equations $y = e^x$ and $y = \ln(x)$ are reflections of each other about the line $y = x$ . This is because $e^{\ln(x)} = \ln(e^{x}) = x$ , and so if $(a,b) = (a,e^a)$ lies on the graph of the equation $y = e^x$ , then the point $(b, \ln(b)) = (b, \ln(e^a)) = (b,a)$ lies on the graph of the equation $y = \ln(x)$ .

Maths; Exponentials and logarithms; KS5 Year 12; Natural logarithms

As $x$ decreases, $e^x$ tends towards $0$ . Consequently, the graph of the equation $y = \ln(x)$ has the $y$ -axis as an asymptote. This means that $\ln(x)$ is only defined when $x$ is a positive real number.

$e^0 = 1$ , and so $\ln(1) = 0$ . The graph of the equation $y = \ln(x)$ therefore passes through the point $(1,0)$ .

Example 1

Solve the equation $\ln(x-5) + \ln(x+2) = \ln(18)$ 

Use the multiplication law for logarithms to get that $\ln(x-5) + \ln(x+2) = \ln((x-5)(x+2))$ . Expand this, and raise $e$ to the power of both sides of the equation:

$\begin{aligned}\ln((x-5)(x+2)) &= \ln(18)\\\ln(x^2 - 3x - 10) &= \ln(18)\\e^{\ln(x^2 - 3x - 10)} &= e^{\ln(18)}\\x^2 - 3x - 10 &= 18\\x^2 - 3x - 28 &= 0\\(x-7)(x+4) &= 0\end{aligned}$ 

It looks like the solutions to the equation are $x = 7$ and $x = -4$ . However, $\ln(x-5)$ is only defined when $x \gt 5$ , and $\ln(x+2)$ is only defined when $x \gt -2$ . So, $\ln(x-5) + \ln(x+2)$ is only defined when $x \gt 5$ .

Therefore, the only solution to the equation $\ln(x-5) + \ln(x+2) = \ln(18)$ is $\underline{x=7}$ .