Example

Example 10.38: Solving 2log5x=log5812\log_5 x = \log_5 81

To solve the logarithmic equation 2log5x=log5812\log_5 x = \log_5 81, first use the Power Property of Logarithms to move the coefficient inside the logarithm as an exponent, yielding log5x2=log581\log_5 x^2 = \log_5 81. Since both sides of the equation now have the same base (55), apply the One-to-One Property of Logarithmic Equations to set the arguments equal: x2=81x^2 = 81. Solving this quadratic equation using the Square Root Property gives x=±9x = \pm 9. Because the argument of a logarithm must be strictly positive, the negative value x=9x = -9 is an extraneous solution and is eliminated. The only valid solution is x=9x = 9. Checking the result by substitution gives 2log59=log592=log5812\log_5 9 = \log_5 9^2 = \log_5 81, which confirms the solution.

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Updated 2026-05-25

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Ch.10 Exponential and Logarithmic Functions - Intermediate Algebra @ OpenStax

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