Question 326073
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2e^{3x}\ =\ 4e^{5x}]


Divide both sides by 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{3x}\ =\ 2e^{5x}]


Take the natural log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{3x}\right)\ =\ \ln\left(2e^{5x}\right)]


The log of the product is the sum of the logs.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{3x}\right)\ =\ \ln(2)\ +\ \ln\left(e^{5x}\right)]


Use the rule: *[tex \Large \log_b(x^n)\ =\ n\log_b(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\,\cdot\,\ln\left(e\right)\ =\ \ln(2)\ +\ 5x\,\cdot\,\ln\left(e\right)]


Use: *[tex \Large \log_b(b)\ =\ 1] 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ =\ \ln(2)\ +\ 5x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x\ =\ \ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\ln(2)}{-2}\ \approx\ -0.347]


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