# Slope lines to y=e^bx

Gold Member
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?

Last edited:

Mark44
Mentor
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 = ebx, which can't be solved by ordinary means, although you can use numerical methods to get a close approximation.
WWGD said:
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?

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

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##.