# Thread: Tangent to the Curve

1. ## Tangent to the Curve

Greetings,

This should be a relatively easy question, but, unfortunately, I am struggling with solving it. I understand the forum is used purely for guiding someone. However, I have the GRE Math Subject test in 13hrs and need to see how this problem is worked out if someone wouldn't mind.

The question is:

For what value of b is the line $\displaystyle y=10x$ tangent to the curve $\displaystyle y=e^{bx}$ at some point in the xy-plane?

Thanks again.

2. Originally Posted by dwsmith
Greetings,

This should be a relatively easy question, but, unfortunately, I am struggling with solving it. I understand the forum is used to purely for guiding someone. However, I have the GRE Math Subject test in 13hrs and need to see how this problem is worked out if someone wouldn't mind.

The question is:

For what value of b is the line $\displaystyle y=10x$ tangent to the curve $\displaystyle y=e^{bx}$ at some point in the xy-plane?

Thanks again.
tangent line "touches" the exponential function ...

$e^{bx} = 10x$

slope of the curve = slope of the tangent line

$be^{bx} = 10$

$\displaystyle \frac{be^{bx}}{e^{bx}} = \frac{10}{10x}$

$\displaystyle b = \frac{1}{x}$

$e^{bx} = 10x$

$e = 10x$

$\displaystyle x = \frac{e}{10}$

line is tangent to the curve at $\displaystyle \left(\frac{e}{10} , e\right)$

$\displaystyle b = \frac{10}{e}$

3. Originally Posted by skeeter
tangent line "touches" the exponential function ...

$\displaystyle \frac{be^{bx}}{e^{bx}} = \frac{10}{10x}$
Why did you divide the derivatives by the original equations?

4. Originally Posted by dwsmith
Why did you divide the derivatives by the original equations?
to get rid of $e^{bx}$