Find the value of A such as that the tangent to y=e^x is a line through origin when x=a
This is what I've done so far,
Derivative of e^x is e^x, so the slope is e^x
(y-e^a)=e^a(x-a)
0-e^a=e^a(0-a)
-e^a=-a(e^a)
That's what I get in the end, how can I solve for a? Did I make an error? This corresponds to a question asking a similar question for lnx, where a would be the value e, so I suppose there has to be some connection....