1Cademy - Try It 10.79 and 10.80: Solving $$\log(x+2) - \log(4x+3) = -\log x$$ and $$\log(x-2)

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Solving Logarithmic Equations by Condensing

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Try It 10.79 and 10.80: Solving $\log(x+2) - \log(4x+3) = -\log x$ and $\log(x-2) - \log(4x+16) = \log \frac{1}{x}$

Practice solving logarithmic equations with logarithms on both sides by first condensing them. For the equation $\log(x+2) - \log(4x+3) = -\log x$ , apply the Quotient Property on the left to get $\log\left(\frac{x+2}{4x+3} ight)$ and the Power Property on the right to get $\log x^{-1}$ , or $\log\left(\frac{1}{x} ight)$ . Using the One-to-One Property of Logarithmic Equations, set the arguments equal: $\frac{x+2}{4x+3} = \frac{1}{x}$ . Cross-multiplying yields $x(x+2) = 4x+3$ , which simplifies to $x^2 - 2x - 3 = 0$ . Factoring gives $(x-3)(x+1) = 0$ , resulting in potential solutions $x = 3$ and $x = -1$ . Because evaluating the original logarithm for $x = -1$ results in a negative argument, it is an extraneous solution, leaving only $x = 3$ . For the equation $\log(x-2) - \log(4x+16) = \log \frac{1}{x}$ , condensing the left side yields $\log\left(\frac{x-2}{4x+16} ight)$ . Equating the arguments gives $\frac{x-2}{4x+16} = \frac{1}{x}$ . Cross-multiplying yields $x^2 - 2x = 4x + 16$ , which simplifies to $x^2 - 6x - 16 = 0$ . Factoring gives $(x-8)(x+2) = 0$ , so the potential solutions are $x = 8$ and $x = -2$ . The value $x = -2$ is extraneous, so the only valid solution is $x = 8$ .

Updated 2026-05-25

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