12 Patterns of Logarithmic Equations

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.

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

YouTube player

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 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 Pythagoras Theorem Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume




Related Articles

Responses

Your email address will not be published. Required fields are marked *