Example

Try It 10.77 and 10.78: Solving log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2(x-2) = 3 and log⁡2x+log⁡2(x−6)=4\log_2 x + \log_2(x-6) = 4

Practice solving logarithmic equations by first condensing them into a single logarithm using the Product Property, then converting to exponential form. For the equation log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2(x-2) = 3, condense the left side to get log⁡2(x(x−2))=3\log_2(x(x-2)) = 3. Convert to exponential form: 23=x2−2x2^3 = x^2 - 2x, which simplifies to 8=x2−2x8 = x^2 - 2x. Setting to zero gives 0=x2−2x−80 = x^2 - 2x - 8. Factoring yields 0=(x−4)(x+2)0 = (x-4)(x+2). The potential solutions are x=4x = 4 and x=−2x = -2. Since a logarithm cannot be evaluated for a negative number, x=−2x = -2 is an extraneous solution, leaving the valid solution x=4x = 4. For the equation log⁡2x+log⁡2(x−6)=4\log_2 x + \log_2(x-6) = 4, condense it to log⁡2(x(x−6))=4\log_2(x(x-6)) = 4. In exponential form, this is 24=x2−6x2^4 = x^2 - 6x, or 16=x2−6x16 = x^2 - 6x. Subtract 16 to get 0=x2−6x−160 = x^2 - 6x - 16. Factoring gives 0=(x−8)(x+2)0 = (x-8)(x+2). The potential solutions are x=8x = 8 and x=−2x = -2. Again, x=−2x = -2 is extraneous, leaving the valid solution x=8x = 8.

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Updated 2026-06-17

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