We can use the laws of logarithms to write equations in different forms. This can be particularly useful if an unknown appears as an index (exponent).
$$ \large 2^x=7$$
For the logarithmic function, for every value of $y$, there is only one corresponding value of $x$.
$$ \large y=5^x$$
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 $\log_{10}{x} + \log_{10}{(x-3)} = \log_{10}{4}$.
\( \begin{align} \displaystyle
\log_{10}{x} + \log_{10}{(x-3)} &= \log_{10}{4} \\
\log_{10}{x(x-3)} &= \log_{10}{4} \\
x(x-3) &= 4 \\
x^2-3x &= 4 \\
x^2-3x-4 &= 0 \\
(x-4)(x+1) &= 0 \\
x &= 4 \text{ or } x = -1 \\
\therefore x &=4 &x>0
\end{align} \)
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