# Thread: Natural Logarithm: find the point on the curve

1. ## Natural Logarithm: find the point on the curve

Question: Find the point on the curve y=e^2x+1 where the tangent is parallel to the line y=2ex.

2. ## Re: Natural Logarithm: find the point on the curve

Originally Posted by onehundredpercenteffort
Question: Find the point on the curve y=e^2x+1 where the tangent is parallel to the line y=2ex.
The graph of $y = e^2x+1$ is a straight line, so a tangent to it is itself. Since the slope of $y = e^2x+1$, which is $e^2$, is different from the slope of $y = 2ex$, which is $2e$, no tangent to $y = e^2x+1$ is parallel to $y = 2ex$.

3. ## Re: Natural Logarithm: find the point on the curve

Originally Posted by onehundredpercenteffort
Question: Find the point on the curve y=e^2x+1 where the tangent is parallel to the line y=2ex.
Is the function $e^{2x}+1\text{ or }e^{2x+1}$ or something else altogether?

4. ## Re: Natural Logarithm: find the point on the curve

Whatever the function, you need to take the derivative and set it equal to 2e.

- Hollywood

5. ## Re: Natural Logarithm: find the point on the curve

i understand that i have to find the derivative which was y'=2e^2x+1 and since the tangent is parallel, the slopes will be the same. But how can i find the point??

6. ## Re: Natural Logarithm: find the point on the curve

Originally Posted by onehundredpercenteffort
i understand that i have to find the derivative which was y'=2e^2x+1 and since the tangent is parallel, the slopes will be the same. But how can i find the point??
Why do you refuse to tell us what the function really is?

What is the correct way to read the function? See reply #3.

7. ## Re: Natural Logarithm: find the point on the curve

You set y' = 2e^(2x+1) = 2e and solve for x. The slope of the other curve, y=2ex, is 2e. So x=0, right?

All three of us were confused about what the function was; you really should have cleared that up in addition to asking for clarification.

- Hollywood