Question 359107
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Your function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ W(t)\ =\ 1000\,\cdot\,2^{-0.04t}]


is only valid if you start with a kilogram of a particular substance.  The correct general formula for that particular substance is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ W(t)\ =\ W_o\,\cdot\,2^{-0.04t}]


Where *[tex \Large W_o] is the starting weight of the substance.


If we are interested in the time for the weight to halve, then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{W(t)}{W_o}\ =\ \frac{1}{2}]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{-0.04t}\ =\ 0.5]


Take the log of both sides (any base is ok, but you should use either base 10 or base e because those are generally what is supported by your calculator)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ln\left(2^{-0.04t}\right)\ =\ \ln(0.5)]


Use the laws of logarithms


*[tex \LARGE \ \ \ \ \ \ \ \ \ -0.04t\ln\left(2\right)\ =\ \ln(0.5)]


Divide by *[tex \Large -0.04\ln(2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ t\ =\ \frac{\ln(0.5)}{-0.04\ln\left(2\right)}]


The rest is calculator work.


Part b is done the same way, except:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{W(t)}{W_o}\ =\ 0.01]


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