Example

Example 10.24: Solving loga49=2\log_a 49 = 2 and lnx=3\ln x = 3

To solve logarithmic equations such as loga49=2\log_a 49 = 2 and lnx=3\ln x = 3, first convert them into their equivalent exponential forms. For the equation loga49=2\log_a 49 = 2, rewriting it in exponential form yields a2=49a^2 = 49. Using the square root property, the potential solutions are a=7a = 7 and a=7a = -7. Because the base of a logarithm cannot be negative, a=7a = -7 is an extraneous solution and must be eliminated, leaving a=7a = 7. Verifying this by substitution confirms that 72=497^2 = 49. For the natural logarithmic equation lnx=3\ln x = 3, rewriting it into exponential form gives x=e3x = e^3. Therefore, the exact solution is x=e3x = e^3. Checking this result by substituting it back into the original equation confirms that lne3=3\ln e^3 = 3.

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Updated 2026-05-26

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

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