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### Logarithms (Economics)

#### Definition

A logarithm is a way of expressing the exponent to which a given number (the base) must be raised to to get another number. Logarithms are often used to solve equations (see below). If we must raise the number $a$ to the power of $b$ to get the number $x$ then we write: $\log_a x=b$ This is equivalent to $a^b=x$.

Logarithms can be used to find the exponent to which a number must be raised to get another number (see below for examples).

The most common bases are $10$ and $e$ (see Exponential Function). Conventionally, $\log$ written without a specific base means $\log_{10}$, and $\ln$ means $\log_ e$. A logarithm to the base $e$ ($\approx 2.71828\dots$) is called the natural logarithm.

For example, we must raise $10$ to the power of $2$ to get $100$ ($10^2=100$) so we can write this as: $\log_{10} 100=2 \text{ or } \log 100=2$

##### Useful Results

Some useful results to remember are \begin{align} \log(10) &= 1, & \ln ( e) &= 1,\\ \log_a (a) &= 1, & \log_a (1) &= 0,\\ \log_a (a ^b) &= b, & a ^{\log_a(b)} &= b. \end{align}

#### Laws of logarithms

\begin{align} \log_a(xy) &= \log_a(x) + \log_a(y) & \textbf{(1)} \\ \log_a\left(\dfrac{x}{y}\right) &= \log_a(x) - \log_a(y) & \textbf{(2)} \\ \log_a(x^b) &= b\log_a(x) & \textbf{(3)} \\ \log_b(a) &= \dfrac{1}{\log_a(b)} & \textbf{(4)} \\ \log_a(x) &= \dfrac{\log_b(x)}{\log_b(a)} & \textbf{(5)} \end{align}

#### Worked Examples

###### Example 1

Simplify $3\log (x) - 4\log (x+3) + \log (y)$.

###### Solution

First, use law 3 to bring the coefficients inside the logarithms. $3\log (x) - 4\log (x+3) + \log (y) = \log (x^3) - \log ((x+3)^4) + \log (y)$ Then combine the three terms using laws 1 and 2. \begin{align} \log (x^3) - \log ((x+3)^4) + \log (y) &=\log (x^3y) - \log( (x+3)^4)\\ &=\log \left(\dfrac{x^3y}{(x+3)^4}\right) \end{align}

###### Example 2

Simplify $2\log (\sqrt{27}) + 2\log (3x) - \dfrac{1}{3} \log (x)$.

###### Solution

Use law 3 to obtain: \begin{align} 2\log (\sqrt{27}) + 2\log (3x) - \dfrac{1}{3} \log (x)&=\log (\sqrt{27}^2) + \log ((3x)^2)- \log ((x)^\frac{1}{3})\\ &= \log (27) + \log (9x^2) - \log (x^\frac{1}{3}) \end{align} Then combine the terms using laws 1 and 2. \begin{align} \log (27) + \log (9x^2) - \log (x^\frac{1}{3}) &= \log(243x^2) - \log(x^\frac{1}{3}) \\ &= \log \left(\dfrac{243x^2}{x^\frac{1}{3} }\right) \\ &= \log \left(243x^{\frac{5}{3} }\right) \end{align} Finally since $243 = 3^{5} = 27^{5/3}$ we can use law 3 to bring out a factor of $\frac{5}{3}$. $\log \left(243x^{\frac{5}{3} }\right) = \frac{5}{3} \log(27x)$

So we have: $2\log (\sqrt{27}) + 2\log (3x) - \dfrac{1}{3} \log (x)=\dfrac{5}{3} \log(27x)$

#### Using Logarithms to Solve Exponential Equations

When asked to solve an equation where the unknown variable appears in the power, we can take the logarithms of both sides and use log law 3 $\left(\log(a^b) = b\log a\right)$ to bring the power down and make it the subject of the equation. We can then use the other log laws to rearrange and solve the equation.

#### Worked Examples

###### Example 1

Suppose we know that $64$ is a power of $2$ but we don't know what that power is. Denote this power by $x$. Use the laws of logarithms to find the value of $x$.

###### Solution

We have: $2^x =64$ By taking the logarithm of both sides of this equation we get: $\log(2^x)=\log(64)$ and using law 3 we can write this as: $x\log(2)=\log(64)$ We can now rearrange and solve for $x$: \begin{align} x&=\dfrac{\log(64)}{\log(2)}\\ &=6 \end{align}

###### Example 2

Solve $3^x=5^{x-2}$.

###### Solution

As the unknown appears in the power we must first take the logarithm of both sides, and then use the log laws to separate the $x$ out.

$3^x=5^{x-2}$ Taking the logarithm of both sides gives: $\log (3^x) = \log (5^{x-2})$ Using law 3 we have: $x\log (3) = (x-2)\log( 5)$ Expand the brackets gives: $x\log( 3) = x\log( 5) - 2\log( 5)$ We must now collect the $x$ terms on one side: \begin{align} x\log( 3) - x\log (5) &= - 2\log( 5)\\ x\log (5) - x\log (3) &= 2\log (5)\\ x(\log (5) - \log (3)) &= 2\log (5) \end{align} Then using law 2 we have: $x\log \left(\frac{5}{3}\right) = 2\log (5)$ By dividing through by $\log \left(\frac{5}{3}\right)$ to get $x$ we can find the value of $x$: $x = \frac{2\log (5)}{\log\left(\frac{5}{3}\right)}$

Note: It is usually better to leave a logarithmic solution in exact algebraic form unless you are asked otherwise.

#### Video Examples

###### Example 1

Prof. Robin Johnson simplifies the expression $\log_{10}(5) + \log_{10}(\sqrt{5})-\log_{10}(25)$.

###### Example 2

Prof. Robin Johnson solves the equation $5^x = 7 \times 3^{1-x}$.

###### Example 3

Prof. Robin Johnson shows how $y^2=3x^3$ can be expressed as a straight line by taking logs.

#### Workbook

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.