5^(y) = 2e^(y) I proceeded to take the natural log of both sides, so yln(5) = ln2e(y) Where do I go from here?
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Originally Posted by benny92000 5^(y) = 2e^(y) I proceeded to take the natural log of both sides, so yln(5) = ln2e(y) Where do I go from here? Use the log addition rule to separate the RHS: $\displaystyle \ln(2e^y) = \ln(2) + \ln(e^y) = \ln(2) + y$
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