Help me solve for b in this equation Here is the equation which I end up to while solving a problem in my Traffic and Transportation Engineering book. 
$$22.05=95.55(e^{-34.8b})-73.5(e^{-94.8b})$$
I will appreciate if you can show me the steps. 
On my own, I tried taking the ln on both side of the equation in order to get raid of e, but I am not getting right result. In fact I am doubting if my approach makes sense. 
My initial approach:
$$\ln(22.05)=[\ln(95.55)+\ln(e^{-34.8b})]-[\ln(73.5)+\ln(e^{-94.8b})]$$
this will eliminate e, and than I solve. but yet I am not getting the right answer.. I doubt this approach makes sense.. so if you have a way out. please show your work. 
 A: As far as I know there are no algebraic techniques that can be used to solve the equation $95.55e^{-34.8b}-73.5e^{-94.8b}=22.05$, so I suspect that one of the following is true:


*

*You have made an error in the work leading up to this point, and are trying to solve the wrong equation

*This really is the right equation to solve, but you are not expected to solve it algebraically.


If not algebraic, how do you solve it?  You could try using software to graph the function $y=95.55e^{-34.8x}-73.5e^{-94.8x}$ and see where it is equal to $22.05$.  See the graph at https://www.desmos.com/calculator/em7dxy3wtx for an example of what this looks like.  The nonzero solution appears to be at approximately $x=0.04007$.
Something else you could do is to introduce a new variable $u=e^{-x}$.  Then expressed in terms of this variable, the equation is
$$22.05=95.55u^{34.8} - 73.5u^{94.8}$$
While this makes the equation look a little bit simpler (in that the variable is no longer in the exponent but is now down at "ground level") it is just as hard to solve; to the best of my knowledge there is no algebraic method that will work.  You could use numerical methods (i.e. software) to approximate the solutions for $u$, and then use $x = -\ln(u)$ to find the values of $x$, but that really doesn't seem to be any better than working with the equation directly.
A: As observed $b=0$ is a solution.  The right side has positive derivative at zero and is zero for large $b$ so there is a positive solution.  You won't find it algebraically, so you need a numeric process.  We can find the maximum of the right side by setting the derivative to zero.
$$95.55\cdot (-34.8)e^{-34.8b}-73.5\cdot(-94.8)e^{-94.8}=0\\
-3215.14+6967.8e^{-60b}=0\\
b=\frac 1{60}\log(\frac{6967.8}{3215.14})\\b\approx 0.01289$$
The positive root will be above this.  You can then do Newton-Raphson with $$f(b)=95.55e^{-34.8b}-73.5e^{-94.8}-22.05\\ f'(b)=95.55\cdot (-34.8)e^{-34.8b}-73.5\cdot(-94.8)e^{-94.8}$$
I find the root at $b \approx 0.040065976$
