Tangent Curve to a Curve

**redsoxfan325** · April 24th 2009, 11:43 PM

I read that for any $a>0, x^a > \ln x$ eventually, though it is possible that $\ln x>x^a$ for finitely many $x$ . I figured that $\ln x = x^a$ has zero or two solutions for any given $a>0$ EXCEPT one. Graphing the equations with different values of $a$ led me to conjecture that this occurs when $a=\frac{1}{e}$ . Maple confirmed that the equation $\sqrt[e]{x} = \ln x$ has one solution: $x=e^e$ . I was wondering two things though:

1) How do you analytically show that if $\sqrt[e]{x} = \ln x$ , then $x=e^e$ ?

2) Is there anything else that's special about the function $\sqrt[e]{x}$ ?

**running-gag** · April 25th 2009, 12:03 AM

Originally Posted by redsoxfan325

I read that for any $a>0, x^a > \ln x$ eventually, though it is possible that $\ln x>x^a$ for finitely many $x$ . I figured that $\ln x = x^a$ has zero or two solutions for any given $a>0$ EXCEPT one. Graphing the equations with different values of $a$ led me to conjecture that this occurs when $a=\frac{1}{e}$ . Maple confirmed that the equation $\sqrt[e]{x} = \ln x$ has one solution: $x=e^e$ . I was wondering two things though:

1) How do you analytically show that if $\sqrt[e]{x} = \ln x$ , then $x=e^e$ ?

2) Is there anything else that's special about the function $\sqrt[e]{x}$ ?

Hi

Let $f(x) = \sqrt[e]{x} - \ln x = x^{\frac{1}{e}} - \ln x$

$f'(x) = \frac{1}{e}\:x^{\frac{1}{e}-1} - \frac{1}{x}$

$f'(x) = \frac{1}{ex}\:\left(x^{\frac{1}{e}} - e\right)$

$f'(x) = 0$ for $x^{\frac{1}{e}} = e$ or $x = e^e$

f is decreasing over $)0 ; e^e]$ and increasing over $[e^e ; +\infty($

And strangely $f(e^e) = 0$

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