+26388 3^x=5x
We can use the Lambert W Function.
The link to the Lambert W Function Calculator is:
http://www.had2know.com/academics/lambert-w-function-calculator.html
\(\begin{array}{|rcll|} \hline 3^x &=& 5x \quad & | \quad \text{change of basis} \\ e^{x\ln(3)} &=& 5x \quad & | \quad : e^{x\ln(3)} \\ 1 &=& 5x \cdot e^{-x\ln(3)} \quad & | \quad : 5\\ \frac{1}{5} &=& x \cdot e^{-x\ln(3)} \quad & | \quad \cdot (- \ln(3)) \\\\ \mathbf{-\frac{\ln(3)}{5}} & \mathbf{=}& \mathbf{-x\ln(3) \cdot e^{-x\ln(3)}} \\ && \text{The Lambert W function is the inverse} \\ && \text{ of the function } f(x) = xe^x \\\\ -x\ln(3) &=& W( -\frac{\ln(3)}{5} ) \quad & | \quad : (- \ln(3)) \\ x &=& \frac{1}{- \ln(3)} \cdot W( -\frac{\ln(3)}{5} ) \\ \hline \end{array} \)
We calculate \(W( -\frac{\ln(3)}{5} ) = W(-0.21972245773)\)
Because the parameter -0.21972245773 is in the interval (-1/e, 0), the funcion returns two values.
\(\begin{array}{|rcll|} \hline v_1 = -2.384291 \\ v_2 = -0.295163 \\ \hline \end{array}\)
We calculate x1 and x2:
\(\begin{array}{|rcll|} \hline x_1 &=& \frac{1}{- \ln(3)} \cdot v_1 \\ x_1 &=& -\frac{1}{\ln(3)} \cdot (-2.384291) \\ \mathbf{x_1} &\mathbf{=} & \mathbf{2.170275} \\\\ x_2 &=& \frac{1}{- \ln(3)} \cdot v_2 \\ x_2 &=& -\frac{1}{\ln(3)} \cdot (-0.295163) \\ \mathbf{x_2} &\mathbf{=} & \mathbf{0.268669} \\ \hline \end{array} \)
The image:
+118658 Hi Arshan, :)
I have only seen this for the first time tonight.
Based on what i watched in this clip
https://www.youtube.com/watch?v=ujujm47KLLI
I think what you have drawn looks fine. I am not sure about the rules for the dummy activity.
Since you are given the Lead time (which I am assuming is the same as task duration)
Perhaps you could include the cross hairs as shown on this video clip
