Question 278941
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12e^{8t}\ -\ e\ =\ 8e^{8t}]


Yes, the first step is to put both of the *[tex \Large e^{8t}] terms on one side.  Just for the sake of what I consider to be neatness, I'm going to add *[tex \Large  -8e^{8t}\ +\ e] to both sides, with the following result:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4e^{8t}\ =\ e]


Next, take the natural log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(4e^{8t}\right)\ =\ \ln(e)]


The log of the product is the sum of the logs, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(4)\ +\ \ln\left(e^{8t}\right)\ =\ \ln(e)]


The log of a base raised to a power is the power times the log of the base:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(4)\ +\ 8t\ln\left(e\right)\ =\ \ln(e)]


Since we know that *[tex \Large \ln(e)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln(4)\ +\ 8t\ =\ 1]


Add *[tex \Large -\ln(4)] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8t\ =\ 1\ -\ ln(4)]


Multiply by *[tex \Large \frac{1}{8}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{1\ -\ ln(4)}{8}\ \approx\ -0.04829]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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