finding point of tangency of y=mx and y=e^x

[wine]

at what value of m is the line y=mx through the origin tangent to y=e^x? what are the coordinate of the point of tangency?
I did
y=e^xy'=e^x
y'=e^0
y'=1
mx=e^x
ln mx=lne^x
lnmx=x
lnm + lnx = x

But i dont know if i did that right and I dont know how I can continue! Thank you

 

[Deleted member 4993]

at what value of m is the line y=mx through the origin tangent to y=e^x? what are the coordinate of the point of tangency?
I did
y=e^x
y'=e^x
y'=e^0 ← Why?? that is incorrect!
y'=1
mx=e^x
ln mx=lne^x
lnmx=x
lnm + lnx = x

But i dont know if i did that right and I dont know how I can continue! Thank you

Let the point of tangency on the curve be (x₁,y₁) or (x₁, \(\displaystyle e^{x_1}\))

we know at the tangent point m = y' = \(\displaystyle e^{x_1}\)

So the equation of the line y = mx becomes

\(\displaystyle e^{x_1}\) = \(\displaystyle e^{x_1}\) * x₁

thus for the tangent point (x₁,y₁) we have x₁ = 1. What is y₁?

Make sure that (x₁,y₁) satisfies both y = e^x and y = mx

 

[wine]

Let the point of tangency on the curve be (x₁,y₁) or (x₁, \(\displaystyle e^{x_1}\))

we know at the tangent point m = y' = \(\displaystyle e^{x_1}\)

So the equation of the line y = mx becomes

\(\displaystyle e^{x_1}\) = \(\displaystyle e^{x_1}\) * x₁

thus for the tangent point (x₁,y₁) we have x₁ = 1. What is y₁?

Make sure that (x₁,y₁) satisfies both y = e^x and y = mx

But why m=e^x1 because y=mx may not be parallel to the y=e^x so shouldn't m may not be e^x1?

 

[Deleted member 4993]

But why m=e^x1 because y=mx may not be parallel to the y=e^x so shouldn't m may not be e^x1?

At x = x₁, y = mx is tangent to the curve y = e^x - hence those are "parallel" at that point (x₁, y₁).

 

[wine]

At x = x₁, y = mx is tangent to the curve y = e^x - hence those are "parallel" at that point (x₁, y₁).

I'm guesting that the question means that y=mx cross the tangent of y=e^x when x=x1?

 

[stapel]

I'm guesting that the question means that y=mx cross the tangent of y=e^x when x=x1?

The exercise referred to the line y1 = e^x and a line y2 = mx which is tangent to it. On what basis have you decided that y2 is no longer tangent to y1, but "crosses the tangent" to y1? Do you know what it means for y2 to be "tangent" to y1?

 

[pka]

at what value of m is the line \(\displaystyle y=mx\) through the origin tangent to \(\displaystyle y=e^x~?\) what are the coordinate of the point of tangentcy?

Having read this entire thread carefully, I can only think that you, wine, have serious misunderstandings about this question. Please make a careful study of this page. That graph is the complete solution to your question. Counter to my use of the Moore-method for learning, in this case I think you might gain from the graph. Please see that there is no reason for your using the word parallel. You must sort out these different terms.

 

[Steven G]

at what value of m is the line y=mx through the origin tangent to y=e^x? what are the coordinate of the point of tangency?
I did
y=e^xy'=e^x
y'=e^0
y'=1
mx=e^x
ln mx=lne^x
lnmx=x
lnm + lnx = x

But i dont know if i did that right and I dont know how I can continue! Thank you

This is what the picture looks like for this problem. Draw the graph y=e^x. Now imagine going to every point on this graph and drawing the tangent line at each point. Now look for the line that crosses the origin. The problem is asking where does this line (the one which crosses the origin) intersect the curve y=e^x.
Now you need to figure out how to do this mathematically.
You were given great hints. Here is how I would do this problem.
Let (x1, e^x1) be any point on the curve y=e^x. Then y'(x1)=e^x1. Then the equation of the tangent line is (y-e^x1)=e^x1(x-x1) or y=(e^x1)x +(e^x1 - (e^x1)x1). If this line crosses the origin then (e^x1 - (e^x1)x1)=0 and x1=1. So y1=e^1=e and the answer is (1,e)