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 $$y = \ln(x)$$​.

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 $$\underline{x=7}$$.