I want to show $\log x$ and $e^x$ never meet without having any graphing calculator.
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6Hint: $e^x>x>\log x.$ – Thomas Andrews Jan 21 '20 at 14:51
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2Please explain your mathematics background so that you receive responses that are appropriate to your skill level. For instance, do you know calculus? This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Jan 21 '20 at 14:52
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1Consider the power series for $e^x$ we have \begin{eqnarray} x<1+x< e^x. \end{eqnarray} Now apply his twice. – Donald Splutterwit Jan 21 '20 at 14:54
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I'm not sure whether you are familiar with calculus, but that would be my approach. I'm also not sure if you're using "log" to mean natural log (that's what I use in my answer) or log base 10. My answer can be adapted to work for log base 10 as well.
First note that $e^1=e$ and $\log(1)=0<e$, so $e^x> \ln(x)$ at $x=1$.
Next, $\frac{d}{dx}e^x=e^x$ and $\frac{d}{dx}\ln(x)=\frac{1}{x}$. Thus $\frac{d}{dx}e^x >e$ for $x>1$, and $\frac{d}{dx}\log(x) < 1<e$ for $x>1$. So the derivative of $e^x$ is greater than the derivative of $\ln(x)$ for $x>1$. Hence, if $x>1$: $$e^x =\int_1^xe^t \ dt +e^1>\int_1^x \frac{1}{x} \ dx = \ln(x).$$
For $x<1$ we have $e^x>0>\ln(x)$.
kccu
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I got another approach as e^ x and ln x are inverse of each other so they are symmetric to y = x and for all x》0 f(x)=e^x>x so its image that is ln x will lie below the line hence they will never meet for X greater than zero and for X less than zero there is no confusion – SIMARJEET Jan 23 '20 at 05:14