# 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}

### 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}