Logarithmic Equations Reducible to Quadratics

Logarithmic Equations Reducible to Quadratic of Math Online Tutoring is based on the basic properties of logarithms such as;
$$ \displaystyle
\log_{a}{b} = \frac{1}{\log_{b}{a}} \\
\log_{a}{b} = \frac{\log_{c}{b}}{\log_{c}{a}} \\
$$

Question 1

Solve \(2 \log_{2}{x} – 9 \log_{x}{2} = 3 \).

\( \begin{aligned} \displaystyle
2 \log_{2}{x} – \frac{9}{\log_{2}{x}} &= 3 \\
2 (\log_{2}{x})^2 – 9 &= 3 \log_{2}{x} \\
2 (\log_{2}{x})^2 – 3 \log_{2}{x} – 9 &= 0 \\
(2 \log_{2}{x} + 3)(\log_{2}{x} – 3) &= 0 \\
\log_{2}{x} = -\frac{3}{2} &\text{ or } \log_{2}{x} = 3 \\
x = 2^{-\frac{3}{2}} &\text{ or } x = 2^3 \\
\therefore x = \frac{1}{\sqrt{2^3}} &\text{ or } x = 8 \\
\end{aligned} \\ \)

Question 2

Solve \( \log{2x} + \log{(x-1)} = \log{(x^2+3)} \).

\( \begin{aligned} \displaystyle \require{color}
\log{2x(x-1)} &= \log{(x^2+3)} &\color{red} \log{A} + \log{B} = \log{AB} \\
2x(x-1) &= x^2 + 3 \\
x^2 – 2x – 3 &= 0 \\
(x-3)(x+1) &= 0 \\
x = 3 &\text{ or } x = -1 \\
\therefore x &= 3 &\color{red} x \gt 0 \text{ from } \log{2x} \\
\end{aligned} \\ \)

Question 3

Solve \( \log_{2}{(x-5)} = \log_{4}{(x-2)} + 1 \).

\( \begin{aligned} \displaystyle
\log_{2}{(x-5)} &= \frac{\log_{2}{(x-2)}}{\log_{2}{4}} + 1 \\
\log_{2}{(x-5)} &= \frac{\log_{2}{(x-2)}}{2} + 1 \\
2 \log_{2}{(x-5)} &= \log_{2}{(x-2)} + 2 \\
\log_{2}{(x-5)^2} &= \log_{2}{(x-2)} + \log_{2}{4} \\
\log_{2}{(x-5)^2} &= \log_{2}{4(x-2)} \\
(x-5)^2 &= 4(x-2) \\
x^2 -10x + 25 &= 4x – 8 \\
x^2 – 14x + 33 &= 0 \\
(x-3)(x-11) &= 0 \\
x = 3 \text{ or } x &= 11 \\
\therefore x &= 11 &\color{red} x \gt 5 \text{ from } \log_{2}{(x-5)} \\
\end{aligned} \\ \)

Question 4

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

\( \begin{aligned} \displaystyle
\frac{3}{\log_{10}{x}} + \log_{10}{x} &= 4 \\
3 + (\log_{10}{x})^2 &= 4 \log_{10}{x} \\
(\log_{10}{x})^2 – 4 \log_{10}{x} +3 \\
(\log_{10}{x} – 1)(\log_{10}{x} – 3) &= 0 \\
\log_{10}{x} = 1 &\text{ or } \log_{10}{x} = 3 \\
x = 10^1 &\text{ or } x = 10^3 \\
\therefore x = 10 &\text{ or } x = 1000 \\
\end{aligned} \\ \)

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