# Finding b such that the line $y=10x$ is tangent to $y=e^{bx}$

I need to find the value $b$ such that the line $y = 10x$ is tangent to the curve $y=e^{bx}$ at some point in the x-y plane. The correct answer is $b = \displaystyle \frac{10}{e}$, however, I have been unsuccessful in deriving the answer myself or of convincing myself that it's correct.

What I did was this: The slope of the line $y = 10x$ is $10$, and the slope of the tangent line to the curve $y = e^{bx}$ is $y^{\prime} = be^{bx}$, so then, I set $10=be^{bx}$, and attempted to solve for $b$.

By simply dividing both sides by $b$, you get something that looks temptingly like $b= \displaystyle \frac{10}{e}$, $b = \displaystyle \frac{10}{e^{bx}}$. However, I don't think that this is sufficient. I have been attempting to solve explicitly for $b$ by using natural logs, but I keep on going in circles back to $10=be^{bx}$. I must be missing something here, so any assistance you could give would be most appreciated.

• hint : the tangent passes through $(0,0)$ Commented Sep 28, 2014 at 15:33
• This might be just about the stupidest question I've ever asked, but how do you know that? I know the graph of $y=e^{bx}$ passes through $(0,1)$, but how do you know the tangent passes through $(0,0)$?
– user100463
Commented Sep 28, 2014 at 15:35
• nvm thats not exactly an hint, just an observation i guess (y=10x passes through origin) Commented Sep 28, 2014 at 15:41

Let $(x, e^{bx})$ be the point of contact of $y = e^{bx}$ and $y = 10x$.

$$10x = e^{bx} \qquad (1)$$

Similarly, using your method, we get

$$10 = be^{bx} \qquad (2)$$

Now you have two equations and two unknowns. (Hint, divide two equations)

• If you divide them, you get x=b?
– user100463
Commented Sep 28, 2014 at 16:05
• Check your calculation again. You should get $bx = 1$. Then substitute that in the second equation. Commented Sep 28, 2014 at 16:06
• Aaah! Gotcha. Thanks.
– user100463
Commented Sep 28, 2014 at 16:09