Example

Example 10.40: Solving log4(x+6)log4(2x+5)=log4x\log_4(x+6) - \log_4(2x+5) = -\log_4 x

To solve a logarithmic equation with logarithms on both sides, such as log4(x+6)log4(2x+5)=log4x\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 log4(x+62x+5)\log_4\left(\frac{x+6}{2x+5}\right). Next, use the Power Property on the right side to rewrite log4x-\log_4 x as log4(x1)\log_4(x^{-1}), which simplifies to log4(1x)\log_4\left(\frac{1}{x}\right). Now, apply the One-to-One Property of Logarithmic Equations: since log4(x+62x+5)=log4(1x)\log_4\left(\frac{x+6}{2x+5}\right) = \log_4\left(\frac{1}{x}\right), it must be true that x+62x+5=1x\frac{x+6}{2x+5} = \frac{1}{x}. Solve this rational equation by multiplying by the common denominator to obtain x(x+6)=2x+5x(x+6) = 2x+5, which expands to x2+6x=2x+5x^2+6x = 2x+5. Rearranging into standard form yields x2+4x5=0x^2+4x-5 = 0. Factoring gives (x+5)(x1)=0(x+5)(x-1) = 0, resulting in potential solutions x=5x = -5 and x=1x = 1. Checking these in the original equation, x=5x = -5 results in taking the logarithm of a negative number (e.g., log4(5)-\log_4(-5)), making it an extraneous solution. The only valid solution is x=1x = 1.

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Updated 2026-05-26

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