12 Patterns of Logarithmic Equations

Solving logarithmic equations is done using properties of logarithmic functions, such as multiplying logs and changing the base and reciprocals of logarithms.

$$ \large \begin{aligned} \displaystyle \large a^x = y \ &\large \Leftrightarrow x = \log_{a}{y} \\
\large \log{a} + \log{b} &= \large \log{(a \times b)} \\
\large \log{a}-\log{b} &= \large \log{(a \div b)} \\
\large \log{a^n} &= \large n \log{a} \\
\large \log_{a}{b} &= \large \frac{\log_{c}{a}}{\log_{c}{b}} \\
\large \log_{a}{b} &= \large \frac{1}{\log_{b}{a}} \\
\large \log_{a}{a} &= \large 1
\end{aligned} $$

Worked on Examples of Logarithmic Equations

Question 1

Solve $ 5-\log_{4}{8} = \log_{4}{x} $.

$ \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\log_{4}{4^5}-\log_{4}{8} &= \log_{4}{x} &\color{red} a = a \times 1 = a \times \log_{b}{b} = \log_{b}{b^a } \\
\log_{4}{1024}-\log_{4}{8} &= \log_{4}{x} \\
\log_{4}{\frac{1024}{8}} &= \log_{4}{x} &\color{red} \log{a}-\log{b} = \log{(a \div b)} \\
\log_{4}{128} &= \log_{4}{x} \\
\therefore x &= 128
\end{aligned} $

Question 2

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

$ \begin{aligned} \require{AMSsymbols} \require{color}
x &\gt 0 \text{ and } x-3 \gt 0 \\
\text{that is } x &\gt 3 \color{red}\cdots (1) \\
\log_{10}{x(x-3)} &= \log_{10}{4} &\color{red} \log{a} + \log{b} = \log{(a \times b)} \\
x(x-3) &= 4 \\
x^2-3x-4 &= 0 \\
(x-4)(x+1) &= 0 \\
x &= 4 \text{ or } -1 \\
\therefore x &= 4 &\color{red} \text{ by } (1)
\end{aligned} $

Question 3

Solve $ \log_{4}{x} + \log_{4}{(x-6)} = 2 $.

$ \require{AMSsymbols} \require{color} \begin{aligned}
x &\gt 0 \text{ and } x-6 \gt 0 \\
\text{that is } x &\gt 6 \color{red}\cdots (1) \\
\log_{4}{x(x-6)} &= \log_{4}{4^2} &\color{red} \log{a} + \log{b} = \log{(a \times b)} \\
&&\color{red} a = a \times 1 = a \log_{b}{b} = \log_{b}{b^a} \\
x(x-6) &= 4^2 \\
x^2-6x &= 16 \\
x^2-6x-16 &= 0 \\
(x-8)(x+2) &= 0 \\
x &= 8 \text{ or } -2 \\
\therefore x &= 8 &\color{red} \text{ by } (1)
\end{aligned} $

Question 4

Solve $ \big(\log_{10}{x}\big)^2-2 \log_{10}{x}-3 = 0 $.

$ \require{AMSsymbols} \require{color} \begin{aligned}
\big( \log_{10}{x}-3 \big)\big( \log_{10}{x} + 1 \big) &= 0 \\
\log_{10}{x} &= 3 \text{ or } \log_{10}{x} = -1 \\
x &= 10^3 \text{ or } x = 10^{-1} &\color{red} \log_{a}{b} = x \rightarrow b = a^x \\
\therefore x &= 1000 \text{ or } \frac{1}{10}
\end{aligned} $

Question 5

Solve $ \big(\log_{3}{x}\big)^2-\log_{3}{x^4} + 3 = 0 $.

$ \require{AMSsymbols} \require{color} \begin{aligned}
\big(\log_{3}{x}\big)^2-4 \log_{3}{x} + 3 &= 0 &\color{red} \log_{a}{b^n} = n \log_{a}{b} \\
\big( \log_{3}{x}-3 \big)\big( \log_{3}{x}-1 \big) &= 0 \\
\log_{3}{x} &= 3 \text{ or } \log_{3}{x} = 1 \\
x &= 3^3 \text{ or } x = 3^1 &\color{red} \log_{a}{b} = x \rightarrow b = a^x \\
\therefore x &= 27 \text{ or } 3
\end{aligned} $

Question 6

Solve $ \big(\log_{2}{x}\big)^2 = \log_{2}{x^4} $.

$ \require{AMSsymbols} \require{color} \begin{aligned}
\big(\log_{2}{x}\big)^2 &= 4 \log_{2}{x} &\color{red} \log_{a}{b^n} = n \log_{a}{b} \\
\big(\log_{2}{x}\big)^2-4 \log_{2}{x} &= 0 \\
\log_{2}{x}\big(\log_{2}{x}-4\big) &= 0 \\
\log_{2}{x} &= 0 \text{ or } \log_{2}{x} = 4 \\
x &= 2^0 \text{ or } 2^4 &\color{red} \log_{a}{b} = x \rightarrow b = a^x \\
\therefore x &= 1 \text{ or } 16
\end{aligned} $

Question 7

Solve $ \displaystyle \frac{\log_{10}{x}}{\log_{10}{2}} = 4 $.

$ \begin{aligned} \require{AMSsymbols} \require{color} \displaystyle
\log_{10}{x} &= 4 \times \log_{10}{2} \\
\log_{10}{x} &= \log_{10}{2^4} &\color{red} n \log_{a}{b} = \log_{a}{b^n} \\
x &= 2^4 \\
\therefore x &= 16 \\
\text{Alternatively, } \\
\log_{2}{x} &= 4 &\color{red} \frac{\log_{b}{x}}{\log_{b}{y}} = \log_{y}{x} \\
x &= 2^4 &\color{red} \log_{a}{b} = x \rightarrow b = a^x \\
\therefore x &= 16
\end{aligned} $

Question 8

Solve $ \log_{10}{x^2} + \log_{10}{8x} = 3. $

$ \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\log_{10}{\big(x^2 \times 8x\big)} &= 3 &\color{red} \log{a} + \log{b} = \log {(a \times b)} \\
\log_{10}{8x^3} &= 3 \\
8 x^3 &= 10^3 &\color{red} \log_{a}{b} = c \leftrightarrow b = a^c \\
8 x^3 &= 1000 \\
x^3 &= 125 \\
\therefore x &= 5
\end{aligned} $

Question 9

Solve $ \log_{4}{x}-\log_{8}{x} = 2 $.

$ \begin{aligned} \require{AMSsymbols} \displaystyle
\frac{\log_{2}{x}}{\log_{2}{4}}-\frac{\log_{2}{x}}{\log_{2}{8}} &= 2 &\color{red} \log_{a}{b} = \frac{\log_{c}{b}}{\log_{c}{a}} \\
\frac{\log_{2}{x}}{2}-\frac{\log_{2}{x}}{3} &= 2 &\color{red} \log_{2}{2^2} = 2 \log_{2}{2} = 2 \\
&&\color{red} \log_{2}{2^3} = 3 \log_{2}{2} = 3 \\
3 \log_{2}{x}-2 \log_{2}{x} &= 12 \\
\log_{2}{x} &= 12 \\
x &= 2^{12} \\
\therefore x &= 4096
\end{aligned} $

Question 10

Solve $ \log_{2}{x}-\log_{x}{4} = 1 $.

$ \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\log_{2}{x}-2\log_{x}{2} &= 1 \\
\log_{2}{x}-\frac{2}{\log_{2}{x}} &= 1 &\color{red} \log_{a}{b} = \frac{1}{\log_{b}{a}} \\
\big(\log_{2}{x}\big)^2-2 &= \log_{2}{x} \\
\big(\log_{2}{x}\big)^2-\log_{2}{x}-2 &= 0 \\
\big(\log_{2}{x}-2\big)\big(\log_{2}{x} + 1\big) &= 0 \\
\log_{2}{x} &= 2 \color{red} \cdots (1) \\
\log_{2}{x} &= -1 \color{red} \cdots (2) \\
x &= 2^2 \color{red} \cdots (1) \\
x &= 4 \\
x &= 2^{-1} \color{red} \cdots (2) \\
x &= \frac{1}{2} \\
\therefore x&= 4 \text{ or } \frac{1}{2}
\end{aligned} $

Question 11

Solve $ x^{\log_{10}{x}} = 1000x^2 $.

$ \begin{aligned} \displaystyle \require{AMSsymbols}\require{color}
\log_{10}{x^{\log_{10}{x}}} &= \log_{10}{1000x^2} \\
\log_{10}{x} \times \log_{10}{x} &= \log_{10}{1000} + \log_{10}{x^2} \\
(\log_{10}{x})^2 &= \log_{10}{10^3} + 2 \log_{10}{x} \\
(\log_{10}{x})^2 &= 3 \log_{10}{10} + 2 \log_{10}{x} \\
(\log_{10}{x})^2 &= 3 + 2 \log_{10}{x} \\
(\log_{10}{x})^2-2 \log_{10}{x}-3 &= 0 \\
(\log_{10}{x}-3)(\log_{10}{x}+1) &= 0 \\
\log_{10}{x} &= 3 &\log_{10}{x} &= -1 \\
x &= 10^3 &\ x &= 10^{-1} \\
\therefore x &= 1000 &\ x &= \frac{1}{10}
\end{aligned} $

Question 12

Solve $ 5^{\log_{10}{x}} \times x^{\log_{10}{5}}-3 (5^{\log_{10}{x}} + x^{\log_{10}{5}}) + 5 = 0 $.

$ \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
5^{\log_{10}{x}} \times 5^{\log_{10}{x}}-3 (5^{\log_{10}{x}} + 5^{\log_{10}{x}}) + 5 &= 0 &\color{red} 5^{\log_{10}{x}} = x ^{\log_{10}{5}} \\
(5^{\log_{10}{x}})^2-6 \times 5^{\log_{10}{x}} + 5 &= 0 \\
(5^{\log_{10}{x}}-5 )(5^{\log_{10}{x}}-1) &= 0 \\
5^{\log_{10}{x}} &= 5 &\ 5^{\log_{10}{x}} &= 1 \\
\log_{10}{x} &= 1 &\ \log_{10}{x} &= 0 \\
\therefore x &= 10 &\ x &= 1
\end{aligned} $

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