Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Slope lines to y=e^bx

  1. Jan 8, 2015 #1

    WWGD

    User Avatar
    Science Advisor
    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: Jan 8, 2015
  2. jcsd
  3. Jan 8, 2015 #2

    Mark44

    Staff: Mentor

    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.
     
  4. Jan 8, 2015 #3

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Thanks. What kind of numerical methods though? Maybe using the Lambert W function or
    something?
     
  5. Jan 8, 2015 #4
    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##.
     
  6. Jan 8, 2015 #5

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Ah, nice, so ## \frac {a}{b}=\mathbb e \implies a= \mathbb e b ## .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook