# Tangent Curve to a Curve

• Apr 24th 2009, 11:43 PM
redsoxfan325
Tangent Curve to a Curve
I read that for any $\displaystyle a>0, x^a > \ln x$ eventually, though it is possible that $\displaystyle \ln x>x^a$ for finitely many $\displaystyle x$. I figured that $\displaystyle \ln x = x^a$ has zero or two solutions for any given $\displaystyle a>0$ EXCEPT one. Graphing the equations with different values of $\displaystyle a$ led me to conjecture that this occurs when $\displaystyle a=\frac{1}{e}$. Maple confirmed that the equation $\displaystyle \sqrt[e]{x} = \ln x$ has one solution: $\displaystyle x=e^e$. I was wondering two things though:

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

2) Is there anything else that's special about the function $\displaystyle \sqrt[e]{x}$?
• Apr 25th 2009, 12:03 AM
running-gag
Quote:

Originally Posted by redsoxfan325
I read that for any $\displaystyle a>0, x^a > \ln x$ eventually, though it is possible that $\displaystyle \ln x>x^a$ for finitely many $\displaystyle x$. I figured that $\displaystyle \ln x = x^a$ has zero or two solutions for any given $\displaystyle a>0$ EXCEPT one. Graphing the equations with different values of $\displaystyle a$ led me to conjecture that this occurs when $\displaystyle a=\frac{1}{e}$. Maple confirmed that the equation $\displaystyle \sqrt[e]{x} = \ln x$ has one solution: $\displaystyle x=e^e$. I was wondering two things though:

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

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

Hi

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

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

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

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

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

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