Resolva-a-equa-o-abaixo-5-x-16-x-1-x-100-

Question Number 128038 by Walt123 last updated on 03/Jan/21

Resolva a equação abaixo:

5^{x} . 16^{\frac{x - 1}{x}} = 100

Answered by MJS_new last updated on 03/Jan/21

5^{x} 16^{1 - \frac{1}{x}} = 100

5^{x} 16^{- \frac{1}{x}} = \frac{25}{4}

x \ln 5 - \frac{1}{x} \ln 16 = \ln \frac{25}{4}

x^{2} - \frac{2 \ln \frac{5}{2}}{\ln 5} x - \frac{4 \ln 2}{\ln 5} = 0

x = - \frac{2 \ln 2}{\ln 5} \lor x = 2

Answered by rodrigue last updated on 03/Jan/21

n e e d h e l p i {d o n}^{'} t k n o w t o c r e a t e n e w s u b j e c t

\lim_{x \to Π} \frac{2 s i n 2 x}{c o s x + 1}

Commented by liberty last updated on 03/Jan/21

\lim_{x \to π} \frac{2 \sin 2 x}{\cos x + 1}; let x = π + t

\lim_{t \to 0} \frac{2 \sin (2 π + 2 t)}{\cos (π + t) + 1} = \lim_{t \to 0} \frac{2 \sin 2 t}{- \cos t + 1}

\lim_{t \to 0} \frac{2 \sin 2 t}{2 \sin^{2} (\frac{t}{2})} = \infty

L^{'} Hopital \Rightarrow \lim_{t \to 0} \frac{4 \cos 2 t}{\sin t} = \infty

Commented by rodrigue last updated on 04/Jan/21

t h a n k s