Example

Try It 10.40: Solving logx81=2\log_x 81 = 2, log3x=5\log_3 x = 5, and log13127=x\log_{\frac{1}{3}} \frac{1}{27} = x

To find the value of xx in the logarithmic equations logx81=2\log_x 81 = 2, log3x=5\log_3 x = 5, and log13127=x\log_{\frac{1}{3}} \frac{1}{27} = x, convert each into its corresponding exponential form. For logx81=2\log_x 81 = 2, the equivalent exponential equation is x2=81x^2 = 81. Solving for xx yields x=9x = 9 or x=9x = -9. Because a logarithmic base must be positive, x=9x = -9 is eliminated, resulting in x=9x = 9. For log3x=5\log_3 x = 5, the exponential form is 35=x3^5 = x, which evaluates to x=243x = 243. For log13127=x\log_{\frac{1}{3}} \frac{1}{27} = x, the exponential equation is \left(\frac{1}{3} ight)^x = \frac{1}{27}. Expressing 127\frac{1}{27} as a power of 13\frac{1}{3} gives \left(\frac{1}{3} ight)^x = \left(\frac{1}{3} ight)^3. With the same base on both sides, the exponents are equal, yielding x=3x = 3.

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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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