Question 273741
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I don't use graphing calculators, but a graph of your function looks like:


{{{drawing(
500, 500, -5, 25, -2, 35,
grid(1),
graph(
500, 500, -5, 25, -2, 35,
0.48*ln(x+1)+27))}}}


Part c:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ 0.48\,\ln\left(x\,+\,1\right)\ +\ 27]


The question:  What is the value of *[tex \Large x] required to make the value of *[tex \Large f(x)] be 28?


You need to solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.48\,\ln\left(x\,+\,1\right)\ +\ 27\ =\ 28]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.48\,\ln\left(x\,+\,1\right)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(x\,+\,1\right)\ =\ \frac{1}{0.48}\ \approx 2.083]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^{\frac{1}{0.48}\ =\ x\ +\ 1]


If your calculator has an *[tex \Large e^x] function (it might be the inverse of  *[tex \Large \ln(x)]), calculate the approximate value of *[tex \Large e^{\frac{1}{0.48}].  It comes out to just a bit more than 8 -- subtract 1 to get a result that is just a bit larger than 7.


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