Solve Slope Line to y=e^bx: Find b Value

[WWGD]

A friend asked me to help him with this; probably simple, but I seem to be missing
something obvious:

For what values of $b$ is the line $y=10x$ tangent to the curve $C(x)=e^{bx}$?

Seems we need $C'(x)=be^{bx} =10$ for some pair $(b,x)$. But this is where it

seems hairy: how do we isolate either $b$ or $10$ ? we can apply $ln$ on both

sides of $C'(x)$ , to get $bx =ln(10/b)=ln10-lnb$ . Then what?

 

[Mark44]

A friend asked me to help him with this; probably simple, but I seem to be missing
something obvious:

For what values of $b$ is the line $y=10x$ tangent to the curve $C(x)=e^{bx}$?

Seems we need $C'(x)=be^{bx} =10$ for some pair $(b,x)$.

For the line to be tangent to the exponential curve, both have to have the same y-value for a given x, and the slope of the tangent to the exponential has to be 10 at that point.

To find the point of tangency, you need to solve 10x = e^bx, which can't be solved by ordinary means, although you can use numerical methods to get a close approximation.

But this is where it

seems hairy: how do we isolate either $b$ or $10$ ? we can apply $ln$ on both

sides of $C'(x)$ , to get $bx =ln(10/b)=ln10-lnb$ . The what?

 

[WWGD]

Thanks. What kind of numerical methods though? Maybe using the Lambert W function or
something?

 

[DarthMatter]

Lets look at $ax = e^{bx}$. Is $ax$ tangential to $e^{bx}$?

$\frac{a}{b} = e^{bx}\Rightarrow x=\frac{\ln(\frac{a}{b})}{b}$. Such an x exists as long as $\frac{a}{b}>0$. However, we also need ${a}\cdot x = e^{bx}\Rightarrow {a}\cdot \frac{\ln(\frac{a}{b})}{b}=\frac{a}{b}$ or $\ln(\frac{a}{b}) = 1$.

 

[WWGD]

Lets look at $ax = e^{bx}$. Is $ax$ tangential to $e^{bx}$?

$\frac{a}{b} = e^{bx}\Rightarrow x=\frac{\ln(\frac{a}{b})}{b}$. Such an x exists as long as $\frac{a}{b}>0$. However, we also need ${a}\cdot x = e^{bx}\Rightarrow {a}\cdot \frac{\ln(\frac{a}{b})}{b}=\frac{a}{b}$ or $\ln(\frac{a}{b}) = 1$.

Ah, nice, so $\frac {a}{b}=\mathbb e \implies a= \mathbb e b$ .