Fórum de Matemática
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number of real roots of the equation \(4^x = x^2\) is

The function \(4^x\) is a strictly increasing function in all its domain.

\(x^2\) strictlydecreases in the set \(]-\infty, 0[\) and increases in the set \(]0,\infty[\)

Therefore, as \(\lim_{x\to -\infty}4^x=0\) and \(\lim_{x\to -\infty}x^2=\infty\)
AND
\(\lim_{x\to 0y}4^x=1\) and \(\lim_{x\to 0}x^2=0\)

with the monotonous condition for negative x in both functions, there is surely and only one zero in the interval \(]-\infty, 0[\)

On the other hand, for x belonging to \(]0, \infty[\), we have

\(4^x=x^2<=>e^{x.ln(4)}=e^{2.ln(x)}\)
So
you can see that the line

\(y=(ln(4)/2).x\) never intersects \(ln(x)\)

_________________
José Sousa
_{se gostou da resposta, divulgue o fórumdematemática.org}

O Binômio de Newton é tão belo como a Vênus de Milo.
O que há é pouca gente para dar por isso.

óóóó---óóóóóó óóó---óóóóóóó óóóóóóóó
(O vento lá fora.)

Álvaro de Campos, 15-1-1928

Thanks Sir