1Cademy - Example 10.40: Solving $$\log_4(x+6)

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Solving Logarithmic Equations by Condensing

Example

Example 10.40: Solving $\log_4(x+6) - \log_4(2x+5) = -\log_4 x$

To solve a logarithmic equation with logarithms on both sides, such as $\log_4(x+6) - \log_4(2x+5) = -\log_4 x$ , condense each side into a single logarithm. First, apply the Quotient Property on the left side to get $\log_4\left(\frac{x+6}{2x+5}\right)$ . Next, use the Power Property on the right side to rewrite $-\log_4 x$ as $\log_4(x^{-1})$ , which simplifies to $\log_4\left(\frac{1}{x}\right)$ . Now, apply the One-to-One Property of Logarithmic Equations: since $\log_4\left(\frac{x+6}{2x+5}\right) = \log_4\left(\frac{1}{x}\right)$ , it must be true that $\frac{x+6}{2x+5} = \frac{1}{x}$ . Solve this rational equation by multiplying by the common denominator to obtain $x(x+6) = 2x+5$ , which expands to $x^2+6x = 2x+5$ . Rearranging into standard form yields $x^2+4x-5 = 0$ . Factoring gives $(x+5)(x-1) = 0$ , resulting in potential solutions $x = -5$ and $x = 1$ . Checking these in the original equation, $x = -5$ results in taking the logarithm of a negative number (e.g., $-\log_4(-5)$ ), making it an extraneous solution. The only valid solution is $x = 1$ .

Updated 2026-06-18

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OpenStax Intermediate Algebra

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