# Logarithmic Equations

We can use the laws of logarithms to write equations in a different form. This can be particularly useful if an unknown appears as an index (exponent).
$$2^x=7$$
For the logarithmic function, for every value of $y$, there is only one corresponding value of $x$.
$$y=5^x$$
We can, therefore, take the logarithm of both sides of an equation without changing the solution. However, we can only do this if both sides are positive.

The equation $\log_{a}{y} = x$ is an example of a general logarithmic equation. Laws of logarithms and exponents (indices) are used to solve these equations.

### Example 1

Write $y=a^3b^2$ as logarithmic equations in base $10$.

\begin{align} \displaystyle \log_{10}{y} &= \log_{10}{(a^3b^2)} \\ \log_{10}{y} &= \log_{10}{a^3} + \log_{10}{b^2} \\ \therefore \log_{10}{y} &= 3\log_{10}{a} + 2\log_{10}{b} \\ \end{align}

### Example 2

Write $\log{x}=\log{a} + 2\log{b} – 3\log{c}$ without logarithms.

\begin{align} \displaystyle \log{x} &= \log{a} + 2\log{b} – 3\log{c} \\ \log{x} &= \log{a} + \log{b^2} – \log{c^3} \\ \log{x} &= \log{\dfrac{ab^2}{c^3}} \\ \therefore x &= \dfrac{ab^2}{c^3} \end{align}

### Example 3

Write $x^2$ the subject of $2\log_{5}{x}=\log_{5}{3a} + 2$.

\begin{align} \displaystyle \log_{5}{x^2} &= \log_{5}{3a} + 2 \times 1 \\ \log_{5}{x^2} &= \log_{5}{3a} + 2 \times \log_{5}{5} \\ \log_{5}{x^2} &= \log_{5}{3a} + \log_{5}{5^2} \\ \log_{5}{x^2} &= \log_{5}{3a} + \log_{5}{25} \\ \log_{5}{x^2} &= \log_{5}{(3a \times 25)} \\ \log_{5}{x^2} &= \log_{5}{75a} \\ \therefore x^2 &= 75a \\ \end{align}

### Example 4

Write $x$ in terms of $y$ given $y=5 \times 2^x$.

\begin{align} \displaystyle \log_{2}{y} &= \log_{2}{(5 \times 2^x)} \\ \log_{2}{y} &= \log_{2}{5} + \log_{2}{2^x} \\ \log_{2}{y} &= \log_{2}{5} + x\log_{2}{2} \\ \log_{2}{y} &= \log_{2}{5} + x \times 1 \\ \log_{2}{y} &= \log_{2}{5} + x \\ x &= \log_{2}{5} – \log_{2}{y} \\ \therefore x &= \log_{2}{\dfrac{5}{y}} \\ \end{align}

### Example 5

Find $x$ if $\log_{3}{9} = x-2$.

\begin{align} \displaystyle \log_{3}{3^2} &= x-2 \\ 2\log_{3}{3} &= x-2 \\ 2 \times 1 &= x-2 \\ 2 &= x-2 \\ x-2 &= 2 \\ \therefore x &= 4 \\ \end{align}

### Example 6

Find $x$ if $\log_{4}{x}=-2$.

\begin{align} \displaystyle \log_{4}{x} &= -2 \\ x &= 4^{-2} \\ x &= \dfrac{1}{4^2} \\ \therefore x &= \dfrac{1}{16} \\ \end{align}

### Example 7

Find $x$ if $3\log_{x}{16}=6$, $x>0$.

\begin{align} \displaystyle \log_{x}{16} &= 2 &\text{divide both sides by 3} \\ x^2 &= 16 \\ x^2 &= 4^2 \\ \therefore x &= 4 \\ \end{align}

### Example 8

Solve $\log_{x}{\dfrac{1}{125}} = -3$.

\begin{align} \displaystyle \log_{x}{\dfrac{1}{125}} &= -3 \\ x^{-3} &= \dfrac{1}{125} \\ x^{-3} &= \dfrac{1}{5^3} \\ x^{-3} &= 5^{-3} \\ \therefore x &= 5 \\ \end{align}

### Example 9

Solve $\log_{10}{x} + \log_{10}{(x-3)} = \log_{10}{4}$.

\begin{align} \displaystyle \log_{10}{x} + \log_{10}{(x-3)} &= \log_{10}{4} \\ \log_{10}{x(x-3)} &= \log_{10}{4} \\ x(x-3) &= 4 \\ x^2 -3x &= 4 \\ x^2 -3x -4 &= 0 \\ (x-4)(x+1) &= 0 \\ x &= 4 \text{ or } x = -1 \\ \therefore x &=4 &x>0 \\ \end{align}