- #1

- 13

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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*.

Please help.

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- Thread starter Morphayne
- Start date

- #1

- 13

- 0

Problem:

Determine the equation of the tangent to the graph of

The answer in the book is

Please help.

- #2

- 1,753

- 1

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

- #3

- 13

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- #4

- 119

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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?

- #5

- 13

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Please. I need this.

- #6

- 119

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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.

- #7

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

- #8

Dick

Science Advisor

Homework Helper

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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?

- #9

- 13

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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?

- #10

- 119

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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?

- #11

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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 ????

- #12

- 119

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

- #13

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Did I get the part with the derivative right?

- #14

- 119

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yes and yes.

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