Resolver: (Lnx)^2-Ln(1/x)=6.... Pergunta de ideia deHardy

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Resolver: (Lnx)^2-Ln(1/x)=6

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$(lnx)^2-ln(\frac{1}{x})=6 \\
\\
(lnx)^2-(-lnx)=6 \\
\\
(lnx)^2+lnx-6=0 \\
\\
\boxed{lnx=y} \\
\\
y^2+y-6=0
\\
\\
S=\{ -3;2 \} \\
\\
lnx=-3 \rightarrow x=\frac{1}{e^3} \\
\\
lnx=2 \rightarrow x =e^2$

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Como resolver: (4-x^2) Ln(2x^2+x)=0

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Hardy April 2020 | 0 Respostas

Resolver: e^2x-8e^x=0

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Como resolver: (4-x^2) Ln(2x^2+x)=0

Resolver: e^2x-8e^x=0

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