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.
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)