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Summary

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>0a>0a>0 other than 111, the range of the function f(x)=axf(x) = a^xf(x)=ax​ is all positive real numbers. Moreover, you have that ax=aya^x = a^yax=ay​ if and only if ax−y=1a^{x-y} = 1ax−y=1​, which in turn happens only if x=yx = yx=y.


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


​log⁡a(b)\log_a(b)loga​(b)​ is defined to be the unique real number xxx​ such that ax=ba^x = bax=b​, i.e:


​alog⁡a(b)=ba^{\log_a(b)} = baloga​(b)=b​​


​For any positive real numbers aaa​ and bbb with a≠1a \ne 1a=1, aaa​ is referred to as the base of the logarithm.


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


log⁡a(ax)=x\log_a(a^x) = xloga​(ax)=x​​


It follows that the functions axa^xax​ and log⁡a(x)\log_a(x)loga​(x)​ are both inverses of each other: alog⁡a(x)=xa^{\log_a(x)} = xaloga​(x)=x​ and log⁡a(ax)=x\log_a(a^x)=xloga​(ax)=x​.


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


Example 1

Rewrite 52=255^2 = 2552=25 using a logarithm.


The equation ax=ba^x = bax=b is equivalent to x=log⁡a(b)x = \log_a(b)x=loga​(b). Substitute a=5a = 5a=5, x=2x = 2x=2 and b=25b = 25b=25.


Therefore, the statement 52=255^2 = 2552=25 can be rewritten as 2=log⁡5(25)‾\underline{2 = \log_5(25)}2=log5​(25)​.


​

The natural logarithm

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


​log⁡e(b)=ln⁡(b)\log_e(b) = \ln(b)loge​(b)=ln(b)​​


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


Example 2

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


5000e−10−9t=50002e−10−9t=12ln⁡(e−10−9t)=ln⁡(12)\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}5000e−10−9te−10−9tln(e−10−9t)​=25000​=21​=ln(21​)​​​


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:

​−10−9t=ln⁡(12)t=−109ln⁡(12)=693,147,180.6…\begin{aligned}-10^{-9}t &= \ln\left(\frac12\right)\\t &= -10^9\ln\left(\frac12\right)\\&= 693,147,180.6\dots\end{aligned}−10−9tt​=ln(21​)=−109ln(21​)=693,147,180.6…​​​




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


​

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FAQs - Frequently Asked Questions

Can logarithms have negative values?

Yes. But you can't insert a negative number into a logarithm.

Why is the logarithm function important?

Given an equation of the form a^x = b, you can simply rewrite it as x = log_a(b) to obtain the solution.

What are logarithms?

The functions a^x and log_a(x) are inverses to each other so logarithms are used to find x in an equation consisting of terms of the form a^x.

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