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^((x−1)/x) =100

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

Answered by MJS_new last updated on 03/Jan/21

5^x 16^(1−(1/x)) =100 5^x 16^(−(1/x)) =((25)/4) xln 5 −(1/x)ln 16 =ln ((25)/4) x^2 −((2ln (5/2))/(ln 5))x−((4ln 2)/(ln 5))=0 x=−((2ln 2)/(ln 5))∨x=2

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

Answered by rodrigue last updated on 03/Jan/21

need help i don′t know to create new subject lim_(x→Π) ((2sin2x)/(cosx +1))

$$ \\ $$$${need}\:{help}\:{i}\:{don}'{t}\:{know}\:{to}\:{create}\:{new}\:{subject} \\ $$$$\underset{{x}\rightarrow\Pi} {\mathrm{lim}}\:\frac{\mathrm{2}{sin}\mathrm{2}{x}}{{cosx}\:+\mathrm{1}} \\ $$

Commented by liberty last updated on 03/Jan/21

lim_(x→π) ((2sin 2x)/(cos x+1)) ; let x=π+t lim_(t→0) ((2sin (2π+2t))/(cos (π+t)+1)) = lim_(t→0) ((2sin 2t)/(−cos t+1)) lim_(t→0) ((2sin 2t)/(2sin^2 ((t/2)))) = ∞ L′Hopital ⇒ lim_(t→0) ((4cos 2t)/(sin t)) = ∞

$$\:\underset{{x}\rightarrow\pi} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\mathrm{2x}}{\mathrm{cos}\:\mathrm{x}+\mathrm{1}}\:;\:\mathrm{let}\:\mathrm{x}=\pi+\mathrm{t} \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\left(\mathrm{2}\pi+\mathrm{2t}\right)}{\mathrm{cos}\:\left(\pi+\mathrm{t}\right)+\mathrm{1}}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\mathrm{2t}}{−\mathrm{cos}\:\mathrm{t}+\mathrm{1}} \\ $$$$\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:\mathrm{2t}}{\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{t}}{\mathrm{2}}\right)}\:=\:\infty\: \\ $$$$\mathrm{L}'\mathrm{Hopital}\:\Rightarrow\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4cos}\:\mathrm{2t}}{\mathrm{sin}\:\mathrm{t}}\:=\:\infty\: \\ $$

Commented by rodrigue last updated on 04/Jan/21

$$ \\ $$$${thanks} \\ $$

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