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In MATLAB, I plotted an exponential distribution (p.d.f.) in time $\lambda e^{-\lambda t}$ but set the $x$ and $y$ axes to logarithmic. The plot looks like below. enter image description here

I'm trying to mathematically find the equation of the curve. I thought setting the $x$ axis to log would obey the transformation $t\rightarrow log(t)$ and setting the $y$ axis to log would just be applying the log to the resulting function. But that was not it - I plotted the resulting function on normal axes to verify.

Can you please help me get the equation of the curve.

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  • $\begingroup$ I am not sure what you mean by the 'equation of the curve'. Aren't you plotting that equation already? $\endgroup$ – caverac Sep 29 '17 at 0:34
  • $\begingroup$ I meant what equation when plotted on linear $x$ and $y$ axes looks like the curve in the figure above, up to a proportionality constant. $\endgroup$ – IanDsouza Oct 9 '17 at 7:10
  • $\begingroup$ I see, in this case @Harry49 already gave you the answer. Just plot $$ f(t) = \log\lambda - \frac{\lambda}{\ln 10}10^t $$ on a linear scale $\endgroup$ – caverac Oct 9 '17 at 11:50
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One has \begin{aligned} \log \left(\lambda e^{-\lambda t}\right) &= \log\lambda + \frac{\ln e^{-\lambda t}}{\ln 10} \\ &= \log\lambda - \frac{\lambda t}{\ln 10} \\ & = \log\lambda -\frac{\lambda}{\ln 10} 10^{\log t} \end{aligned} The dependence of $\log \left(\lambda e^{-\lambda t}\right)$ with respect to $\log t$ is exponential. However, $\log \left(\lambda e^{-\lambda t}\right)$ is linear with respect to $t$.

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