Logarithms

In a nutshell

Exponential functions have inverses, called logarithms. Any equation with an exponent on one side can be rewritten using logarithms.

Inverses of exponential functions

Recall that for any constant $$a>0$$ other than $$1$$, the range of the function $$f(x) = a^x$$​ is all positive real numbers. Moreover, you have that $$a^x = a^y$$​ if and only if $$a^{x-y} = 1$$​, which in turn happens only if $$x = y$$.

Putting these facts together, you get that for any positive real number $$b$$​​, there exists a unique real number $$x$$​ such that $$a^x = b$$. Logarithms are defined with this in mind:

​$$\log_a(b)$$​ is defined to be the unique real number $$x$$​ such that $$a^x = b$$​, i.e:

​$$a^{\log_a(b)} = b$$​​

​For any positive real numbers $$a$$​ and $$b$$ with $$a \ne 1$$, $$a$$​ is referred to as the base of the logarithm.

Given an equation of the form $$a^x = b$$​, you can simply rewrite it as $$x = \log_a(b)$$ to obtain the solution. In other words:

$$\log_a(a^x) = x$$​​

It follows that the functions $$a^x$$​ and $$\log_a(x)$$​ are both inverses of each other: $$a^{\log_a(x)} = x$$​ and $$\log_a(a^x)=x$$​.

The exact real values of logarithms can be found using the button on your calculator.

Example 1

Rewrite $$5^2 = 25$$ using a logarithm.

The equation $$a^x = b$$ is equivalent to $$x = \log_a(b)$$. Substitute $$a = 5$$, $$x = 2$$ and $$b = 25$$.

Therefore, the statement $$5^2 = 25$$ can be rewritten as $$\underline{2 = \log_5(25)}$$.

​

The natural logarithm

The logarithm with the base $$e$$​ is referred to as the natural logarithm due to how commonly used it is, and is written as $$\ln$$. In particular:

​$$\log_e(b) = \ln(b)$$​​

For any positive real number $$b$$​. It has a separate button on your calculator.

Example 2

The number of particles in a sample of a radioactive element $$t$$​ years after initial measurement is modelled by the equation $$N = 5000e^{-10^{-9}t}$$. What is the half-life of the substance to $$3\space s.f.$$​​?

The half-life is the time it takes for the number of particles to fall to half of the initial value. So, rearrange the following equation using the natural logarithm:

$$\begin{aligned}5000e^{-10^{-9}t} &= \frac{5000}{2}\\e^{-10^{-9}t} &= \frac12\\\ln(e^{-10^{-9}t}) &= \ln\left(\frac12\right)\end{aligned}$$​​

Note: The natural logarithm has been taken of both sides in this step. This is allowed since the same operation has taken place on both sides of the equation. Continuing gives:

​$$\begin{aligned}-10^{-9}t &= \ln\left(\frac12\right)\\t &= -10^9\ln\left(\frac12\right)\\&= 693,147,180.6\dots\end{aligned}$$​​

Therefore, the half-life of the substance to $$3\space s.f.$$ is approximately equal to $$\underline{693,000,000}$$ years.

​