Tangent to
y = e^x log(2, x) at the point (1, 0)

Question

Tangent to 
y = e^x log(2, x) at the point (1, 0)

anonymous · Answer

$$e^x(\log_{2}x) $$

zepdrix · Answer

Mmm so what's the problem? Having trouble taking the derivative of a log that isn't base e?

anonymous · Answer

No, I get this as the tangent line equation $$\frac{ ex }{ \log2 }-\frac{ e }{ \log2 }$$ but apparently my math program considers it incorrect

zepdrix · Answer

Hmm ok let's see what's going on.
$$\Large y=e^x\frac{\ln x}{\ln 2}$$
$$\Large y'=\frac{1}{\ln2}\left(e^x \ln x+e^x\frac{1}{x}\right)$$

zepdrix · Answer

Evaluating the function at x=1 gives us,$$\Large y'(1)=\frac{1}{\ln2}\left(\cancel{e^1\ln1}+e\right)$$

Simplifies nicely to give us the slope of our tangent line,$$\Large m=\frac{e}{\ln2}$$

zepdrix · Answer

$$\Large y_{\tan} \quad=\quad mx+b \qquad\to\qquad y_{\tan}\quad=\quad \frac{e}{\ln2}x+b$$

zepdrix · Answer

Plugging in our point gives ussssssssss,
Hmm ya it looks like you did everything correctly :\ strange...

anonymous · Answer

yeah, that's where I'm super confused....

Maybe you can help me out with another problem that I seem to be doing correctly as well, yet keep getting told I'm wrong:

If $20,000 is invested in a savings account offering 3.5% per year, compounded continuously, how fast is the balance growing after 5 years? (Round your answer to the nearest cent.)

I'm getting $764.99

anonymous · Answer

By the looks of it, it seems like a simple pert formula where the initial investment is subtracted and then the outcome is divided by 5 years to find the rate at that year, but apparently I am wrong?

zepdrix · Answer

I think what we're doing is looking for the slope of the line tangent to the function A=Pe^rt at t=5.

So we want $\Large A'(5)$

What you described would be: The average rate of change across those 5 years.
They're looking for the `instanteous rate of change` AT t=5 years.
At least I think that's what they want, based on the wording.

anonymous · Answer

ok, I figured as much, but with having values for all the variable, show would I go about finding the derivative? Or do I find the derivative based on t and leave t empty so there is a variable?

zepdrix · Answer

Correct, we want t `to vary` as we look for the derivative.
We can't plug in t=5 and then take a derivative, that would mean we were holding t constant.

Derivative, then plug 5. :)

anonymous · Answer

ok after derivatizing (coining this word) I get

$$700e ^{.035t}$$

Does that look correct?

zepdrix · Answer

for A'? ya looks good so far!
Now we're allowed to plug in 5.

anonymous · Answer

833.87 was right! Thanks for your help! Though we weren't able to understand why I was getting a no for the first question I'm glad you were able to help me out with my second!

zepdrix · Answer

yay team \c:/

for the first one, perhaps they want you to round to a decimal for each value?

anonymous · Answer

I'll try that