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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.

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  • $\begingroup$ Please use Mathjax to format math text. This link may be useful: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Dave Jun 6 '17 at 23:45
  • $\begingroup$ ln does not "distribute". You can take the ln of both sides of an equation, but you can't take ln of both terms on the right-hand side of the equation. $\endgroup$ – mweiss Jun 6 '17 at 23:45
  • $\begingroup$ Thank you.. the equations are formatted for better viewing now.. I am still waiting for someone to attempt my problem.. $\endgroup$ – Christian Weah Jun 6 '17 at 23:54
  • $\begingroup$ Hint: try $b=0$. $\endgroup$ – Student Jun 6 '17 at 23:54
  • $\begingroup$ I'm not sure if you noticed or not, but $b=0$ is (by inspection) a solution. Do you just need a solution, or all solutions? $\endgroup$ – mweiss Jun 6 '17 at 23:54
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Using whole numbers, consider the equation to be $$f(b)=7350 e^{-474 b/5}-9555 e^{-174 b/5}+2205$$ and rewrite it as $$f(b)=-9555 e^{-174 b/5}\left(1-\frac{10 }{13}e^{-60 b}\right)+2205$$ In a first step, assume that the term in parentheses is close to $1$ and then solve $$-9555 e^{-174 b/5}+2205=0\implies b=\frac{5}{174} \log \left(\frac{13}{3}\right)\approx 0.0421361$$ Just check now the value of the term in parentheses; it is $$1-\frac{30}{169} \left(\frac{3}{13}\right)^{21/29}\approx 0.938611$$ So, the hypothesis was not totally stupid and we have a good starting value for Newton method. The successive iterates will then be $$\left( \begin{array}{cc} n & b_n \\ 0 & 0.04213612267 \\ 1 & 0.04001784112 \\ 2 & 0.04006595215 \\ 3 & 0.04006597619 \end{array} \right)$$ which is the solution for ten significant figures.

You could also solve the problem generating the sequence $$b_n=\frac{5}{174} \log \left(\frac{13}{3} \left(1-\frac{10 e^{-60 b_{n-1}}}{13}\right)\right)\qquad \text{with}\qquad b_1=\frac{5}{174} \log \left(\frac{13}{3}\right)$$ which will be slower than Newton but will converge to the solution $$\left( \begin{array}{cc} n & b_n \\ 1 & 0.04213612267 \\ 2 & 0.04031561412 \\ 3 & 0.04009787032 \\ 4 & 0.04007007968 \\ 5 & 0.04006650462 \\ 6 & 0.04006604424 \\ 7 & 0.04006598495 \\ 8 & 0.04006597732 \\ 9 & 0.04006597633 \\ 10 & 0.04006597621 \\ 11 & 0.04006597619 \end{array} \right)$$

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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.

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  • $\begingroup$ mweiss, Thanks alot. Actually the steps leading to my equation is correct. $\endgroup$ – Christian Weah Jun 7 '17 at 0:17
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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 $b \approx 0.040065976$

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