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 \( 5^{\log_{10}{x}} \times x^{\log_{10}{5}}-3 (5^{\log_{10}{x}} + x^{\log_{10}{5}}) + 5 = 0 \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
5^{\log_{10}{x}} \times 5^{\log_{10}{x}}-3 (5^{\log_{10}{x}} + 5^{\log_{10}{x}}) + 5 &= 0 &\color{red} 5^{\log_{10}{x}} = x ^{\log_{10}{5}} \\
(5^{\log_{10}{x}})^2-6 \times 5^{\log_{10}{x}} + 5 &= 0 \\
(5^{\log_{10}{x}}-5 )(5^{\log_{10}{x}}-1) &= 0 \\
5^{\log_{10}{x}} &= 5 &\ 5^{\log_{10}{x}} &= 1 \\
\log_{10}{x} &= 1 &\ \log_{10}{x} &= 0 \\
\therefore x &= 10 &\ x &= 1
\end{aligned} \)
Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume
