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

$$$$ Resolva a equação abaixo:\n$$$$ $$5^{\mathrm{x}}.{16}^{\frac{\mathrm{x}-1}{\mathrm{x}}}=100$$

Answered by MJS_new last updated on 03/Jan/21

$$5^x{16}^{1-\frac{1}{x}}=100$$ $$5^x{16}^{-\frac{1}{x}}=\frac{25}{4}$$ $$x\ln 5-\frac{1}{x}\ln 16=\ln \frac{25}{4}$$ $$x^2-\frac{2\ln \frac{5}{2}}{\ln 5}x-\frac{4\ln 2}{\ln 5}=0$$ $$x=-\frac{2\ln 2}{\ln 5}\vee x=2$$

Answered by rodrigue last updated on 03/Jan/21

$$$$ $$needhelpi{don}^{\prime }tknowtocreatenewsubject$$ $$\underset{x\to \mathrm{\Pi }}{\lim }\frac{2sin2x}{cosx+1}$$

Commented byliberty last updated on 03/Jan/21

$$\underset{x\to \pi }{\lim }\frac{2\sin 2\mathrm{x}}{\cos \mathrm{x}+1};\mathrm{let}\mathrm{x}=\pi +\mathrm{t}$$ $$\underset{\mathrm{t}\to 0}{\lim }\frac{2\sin (2\pi +2\mathrm{t})}{\cos (\pi +\mathrm{t})+1}=\underset{\mathrm{t}\to 0}{\lim }\frac{2\sin 2\mathrm{t}}{-\cos \mathrm{t}+1}$$ $$\underset{\mathrm{t}\to 0}{\lim }\frac{2\sin 2\mathrm{t}}{2\sin {}^2\left(\frac{\mathrm{t}}{2}\right)}=\mathrm{\infty }$$ $${\mathrm{L}}^{\prime }\mathrm{Hopital}\Rightarrow \underset{\mathrm{t}\to 0}{\lim }\frac{4\cos 2\mathrm{t}}{\sin \mathrm{t}}=\mathrm{\infty }$$

Commented byrodrigue last updated on 04/Jan/21

$$$$ $$thanks$$

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