# Thread: Tangent to e^x through origin

1. ## Tangent to e^x through origin

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....

2. ## Here is what you do

Originally Posted by theowne
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....
$-e^{a}=-a\cdot{e^{a}}$... $a=1$?...o you made a mistake I think you should have $y-0=e^{a}(x-0)$...so $y=e^{a}x$....a=0...you just forgot that the slope for tangent line must be the same point as is shown in the equation for a tangent line $y-f(x_1)=f'(x_1)(x-x_1)$

3. That's what I figured, but what connection exists between that and the value e? The question seems to imply a relation between the answer here, and e..

4. ## The seeming connection would be

Originally Posted by theowne
That's what I figured, but what connection exists between that and the value e? The question seems to imply a relation between the answer here, and e..
$f(x)=\ln(x)$ both give the equation $y=x$? since $y-e=\ln(e)(x-e)$...so $y=\ln(e)x-\ln(e)e+e=x$...andy maybe that $\ln(e)=1,e^{0}=1$?

5. I think my earlier form was correct....(which results in a=1)..I plugged it into graphmatica and I can see the relationship more clearly now...for ln(x), it's when x=e, for e^x, it's when y=e. Thanks.....

6. ## Hmm

Originally Posted by theowne
I think my earlier form was correct....(which results in a=1)..I plugged it into graphmatica and I can see the relationship more clearly now...for ln(x), it's when x=e, for e^x, it's when y=e. Thanks.....
Maybe I am misunderstanding your question then...but based on you think what you just said...maybe you are supposed to realize that since $\ln(x)$ and $e^{x}$ are inverse functions their domain and ranges are switched