A-level Mathematics/CIE/Pure Mathematics 2/Logarithmic and Exponential Functions

Logarithms and Exponents

A logarithm is the inverse function of an exponent.

e.g. The inverse of the function $f(x)=3^x$ is $f^{-1}(x)={\log }_{3}x$.

In general, $y=b^x⟺x={\log }_{b}y$, given that $b>0$.

Laws of Logarithms

The laws of logarithms can be derived from the laws of exponentiation:

$\begin{array}{cc}{x}^{a+b}=x^a\times x^b& ⟺\log a+\log b=\log ab\\ x^{a-b}=x^a÷x^b& ⟺\log a-\log b=\log a/b\\ (x^a{)}^b=x^{ab}& ⟺\log a^b=b\log a\end{array}$

These laws apply to logarithms of any given base

Natural Logarithms

The natural logarithm is a logarithm with base $e$, where $e$ is a constant such that the function $e^x$ is its own derivative.

The natural logarithm has a special symbol: $\ln x$

The graph $y=e^{kx}$ exhibits exponential growth when $k>0$ and exponential decay when $k<0$. The inverse graph is $y=\frac{1}{k}\ln x$. Here is an interactive graph which shows the two functions as inverses of one another.

Solving Logarithmic and Exponential Equations

An exponential equation is an equation in which one or more of the terms is an exponential function. e.g. $5^x=2^{x+2}$. Exponential equations can be solved with logarithms.

e.g. Solve $3^{x+1}=4^{2x-1}$

$\begin{array}{cc}{3}^{x+1}& =4^{2x-1}\\ (x+1)\ln 3& =(2x-1)\ln 4\\ x\ln 3+\ln 3& =2x\ln 4-\ln 4\\ \ln 3+\ln 4& =x(2\ln 4-\ln 3)\\ x& =\frac{\ln 3+\ln 4}{2\ln 4-\ln 3}\\ x& \approx 1.4844\end{array}$

A logarithmic equation is an equation wherein one or more of the terms is a logarithm.

e.g. Solve $\lg x+\lg (x+2)=2$

$\begin{array}{cc}\lg x+\lg (x+2)& =2\\ \lg (x(x+2))& =2\\ x(x+2)& =100\\ x^2+2x& =100\\ (x+1{)}^2& =101\\ x+1& =\sqrt{101}\\ x& =-1\pm \sqrt{101}\end{array}$

Converting Relationships to a Linear Form

In maths and science, it is easier to deal with linear relationships than non-linear relationships. Logarithms can be used to convert some non-linear relationships into linear relationships.

Exponential Relationships

An exponential relationship is of the form $y=a{b}^x$. If we take the natural logarithm of both sides, we get $\ln y=\ln a+x\ln b$. We now have a linear relationship between $\ln y$ and $x$.

e.g. The following data is related with an exponential relationship. Determine this exponential relationship, then convert it to linear form.

x	y
0	5
2	45
4	405

$\begin{array}{cc}\text{Exponential relationship }⟹y& =a{b}^x\\ 5& =a{b}^0=a(1)\\ a& =5\\ y& =5b^x\\ 45& =5b^2\\ 9& =b^2\\ b& =3\\ y& =5(3^x)\end{array}$

Now convert it to linear form by taking the natural logarithm of both sides:

$\begin{array}{cc}y& =5(3^x)\\ \ln y& =\ln 5+x\ln 3\end{array}$

Power Relationships

A power relationship is of the form $y=a{x}^b$. If we take the natural logarithm of both sides, we get $\ln y=\ln a+b\ln x$. This is a linear relationship between $\ln y$ and $\ln x$.

e.g. The amount of time that a planet takes to travel around the sun (its orbital period) and its distance from the sun are related by a power law. Use the following data to deduce this power law:

Planet	Distance from Sun /106 km	Orbital Period /days
Earth	149.6	365.2
Mars	227.9	687.0
Jupiter	778.6	4331

$\begin{array}{cc}\text{Power law}⟹T& =a{R}^b\\ \text{Use Earth data}⟹365.2& =a(149.6^b)\\ \ln 365.2& =\ln a+b\ln 149.6\\ \text{Use Mars data}⟹687.0& =a(227.9^b)\\ \ln 687.0& =\ln a+b\ln 227.9\\ \ln 687.0-\ln 365.2& =\ln a-\ln a+b\ln 227.9-b\ln 149.6\\ \ln \frac{687.0}{365.2}& =0+b(\ln \frac{227.9}{149.6})\\ b& =\frac{\ln \frac{687.0}{365.2}}{\ln \frac{227.9}{149.6}}\approx 1.5011\\ \ln 365.2& =\ln a+1.5011\ln 149.6\\ \ln a& =\ln 365.2-\ln 1839.9\\ \ln a& =\ln 0.1985\\ a& =0.1985\\ \therefore T& =0.1985R^{1.5011}\end{array}$

References

Notes