Question 629771
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{-0.98t}\ =\ 0.07]


Take the natural log of both sides.  You can actually use the log to any base, but since this problem has *[tex \LARGE e] in it, the arithmetic is simpler if you use natural logs.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(e^{-0.98t}\right)\ =\ \ln\left(0.07\right)] 


Use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -0.98t\ln\left(e\right)\ =\ \ln\left(0.07\right)]


Then use


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(b)\ =\ 1]


to write


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -0.98t\ =\ \ln\left(0.07\right)] 


Multiply both sides by *[tex \LARGE \frac{1}{-0.98}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln\left(0.07\right)}{-0.98}]


The rest is just calculator work.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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