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We have to find minimum distance between $e^x$ and $ln x$

I thought they are mirror image along $y=x$ .

So the point which would be at minimum distance would have slope -1 .

from that I got the answer as $\sqrt 2$ . which is correct .

But I want to kno2 ifbthere is some other good method .

a25bedc5-3d09-41b8-82fb-ea6c353d75ae
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  • yes, it is correct, distance from $(0,1)$ on $e^x$ graph to $(1,0)$ on $\ln x$ graph. – Mirko Apr 29 '17 at 17:31
  • @Mirko I want to if there is any other method to solve it – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Apr 29 '17 at 17:42
  • you could minimize the square $d^2$ of the distance between a point $(x,e^x)$ on the graph of $e^x$ and a point $(t,\ln(t))$ on the graph of $\ln(t)$. You get $d^2=f(x,t)=(x-t)^2+(e^x-\ln(t))^2$. Taking partials to find critical points $\frac{\partial f}{\partial x}=2(x-t)+2(e^x-\ln(t))e^x=0$ and $\frac{\partial f}{\partial t}=-2(x-t)-2\frac{(e^x-\ln(t))}t=0$, from which $e^x=\frac1t$ thus $x=-\ln(t)$, one solution clearly $(x,t)=(0,1)$. Substituting back $-\ln(t)-t+\frac1{t^2}-\frac{\ln(t)}t=0$, derivative of left side is negative, so the left side is strictly decreasing, no other solutions. – Mirko Apr 30 '17 at 14:53
  • though for the "no other solutions part" you need to take care of the possibility $e^x=\ln(t)$ but that implies $x=t$ (from the system of the partials), that is $e^t=\ln(t)$, $e^{e^t}=t$, so indeed no other solutions. Seems too late to post this as an answer as they closed this question as presumably a duplicate, but now it seems it was not a duplicate, as it asks for other methods. @AlexM. – Mirko Apr 30 '17 at 15:05

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