OpenStudy (anonymous):

Convert f(x)=76.03e^-0.7x to logarithmic form and describe the graph.

3 years ago
OpenStudy (anonymous):

o y=76.03e^-0.7x o y/76.03 = e^-0.7x o ln y/76.03 = ln e^-0.7x o ln y – ln 76.03 = ln e^-0.7x o ln y = ln e^-0.7x + 4.33 o ln y = -0.7x + 4.33 This is what I've done so far, I don't know if this is right or if I should continue or if I need to start over.

3 years ago
OpenStudy (anonymous):

@jdoe0001 @whpalmer4

3 years ago
OpenStudy (anonymous):

@HelpBlahBlahBlah

3 years ago
OpenStudy (anonymous):

@jdoe @whpalmer4 @HelpBlahBlahBlah I just need you to check if I am right or correct me if I'm wrong and assist in the graphing process.

3 years ago
OpenStudy (anonymous):

@whpalmer4 please help!!!

3 years ago
OpenStudy (whpalmer4):

Yes, if you plot \(\ln y \) vs. \(x\) you'll get a straight line.

3 years ago
OpenStudy (anonymous):

Thank you

3 years ago
OpenStudy (anonymous):

Is it diagonal or straight up and down?

3 years ago
OpenStudy (whpalmer4):

It's a function, so it can't be straight up and down. Beyond that, the way we always worked these was to pick a few points and grind through the numbers. I hope as a first step that it is clear that any function of the form \[e^{-kx}\] is going to be something that decreases as \(x\) increases...

3 years ago
OpenStudy (whpalmer4):

Remember \[e^{-kx} = \frac{1}{e^{kx}}\]

3 years ago
OpenStudy (anonymous):

Yes, I know it is a decaying exponential function.

3 years ago
OpenStudy (whpalmer4):

Here's a plot (without axes) of the natural log of the expression \(76.03e^{-0.7x}\) and of your line equation. The fact that you can only see one line here should tell you something about your solution :-)

3 years ago
OpenStudy (anonymous):

Thank you so much!

3 years ago