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Try It 10.83 and 10.84: Solving 2ex−2=182e^{x-2} = 18 and 5e2x=255e^{2x} = 25

Practice solving exponential equations with base ee by using the natural logarithm. For the equation 2ex−2=182e^{x-2} = 18, first isolate the exponential by dividing by 22 to get ex−2=9e^{x-2} = 9. Take the natural logarithm of both sides, resulting in ln⁡ex−2=ln⁡9\ln e^{x-2} = \ln 9. Using the Power Property and knowing ln⁡e=1\ln e = 1, this simplifies to x−2=ln⁡9x-2 = \ln 9. Adding 22 gives the exact solution x=ln⁡9+2x = \ln 9 + 2, approximately 4.197. For the equation 5e2x=255e^{2x} = 25, divide by 55 to isolate the exponential, yielding e2x=5e^{2x} = 5. Take the natural logarithm of both sides to get ln⁡e2x=ln⁡5\ln e^{2x} = \ln 5, which simplifies to 2x=ln⁡52x = \ln 5. Dividing by 22 gives the exact solution x=ln⁡52x = \frac{\ln 5}{2}, approximately 0.805.

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Updated 2026-06-03

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

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