# Equation of a tangent, with a slope of 'e'.

Help my guys. I'm so hopelessly lost on this question...

Problem:

Determine the equation of the tangent to the graph of y=ln x that has slope e.

The answer in the book is y=ex-2.

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Where is your work? Show some and I will help.

Well I just wrote y=xe as my answer but when I checked my answer it was wrong. I'm guessing that I need to do something using the derivative of y=ln x, but that is as far as know. I'm pretty weak with all this work involving Euler's constant.

exk
y=ex can't be the answer. The general equation of a line is y=mx+b. You are given that the slope is e. What else do you know about the graph of the function and the line tangent to it at some point.

If you think that you need to use the derivative of the function then try using it.

and the last hint:

What does ln(e) equal?

I don't understand. Do I take the answer of y=ln e and plug it into y=ln x? If so how would I solve for x? Can someone please help me with a solution? I have an exam in 4 hours and there will be at least 3 questions like this worth(all together) 30 marks out of a possible 70. I'm in panic mode. Please. I need this. exk
This question has almost nothing to do with "e", it has to do with knowing how to find the equation of the line tangent to the graph. Unless you understand how to get the solution it won't help you when the question is phrased differently on the exam.

here is the last hint:
if the line is tangent to the curve then it shares the x,y coordinates as well as the slope.

you shouldn't need anymore than that.

But I don't have any x,y coordinates. If I did, I wouldn't be stuck.

Dick
Homework Helper
But I don't have any x,y coordinates. If I did, I wouldn't be stuck.
You haven't figured out for what value of x that the slope of the tangent is e. That's why you don't have coordinates. Start by doing that. What's the slope of the tangent to the curve y=ln(x)? For what value of x is that equal to e?

I don't know how to solve it:

y = ln x

-sub. y = e:

e = ln x

Now, here is where I'm stuck. I solve for x right? But how do I do that?

exk
y doesn't equal e.

You said initially that you need to use the derivative of your function, try using it.

Do you remember what the derivative of a function represents?

So;

y = ln x

y' = 1/x

The derivative equation gives us the slope of any tangent, at any point along the given function. What do I do from here?

Should I sub. y' = e?

y' = 1/x

e = 1/x

xe = 1

x = 1/e ????

exk
why can't x = 1/e? (in fact it is)

suppose x = 1/e is correct, how do you get the y coordinate? and finally how do you get the equation of the line?

To get the y-coordinate, I must sub. x = 1/e in to y = ln x. Finally, to get the equation of the tangent I would use y = m(x-p)+q and sub. in the appropriate values. Right?

Did I get the part with the derivative right?

exk
yes and yes.