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Solución de sistemas de ecuaciones/ xy=40; x^lgy=4

xy=40; x^lgy=4

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Solución

Ha introducido [src] 
x*y = 40
xy=40x y = 40 
 log(y)    
x       = 4
xlog⁡(y)=4x^{\log{\left(y \right)}} = 4 
x^log(y) = 4
Respuesta rápida
x1=210e−log⁡(256)+log⁡(40)22x_{1} = \frac{2 \sqrt{10}}{e^{\frac{\sqrt{- \log{\left(256 \right)} + \log{\left(40 \right)}^{2}}}{2}}}
=
210e−log⁡(256)+log⁡(40)22\frac{2 \sqrt{10}}{e^{\frac{\sqrt{- \log{\left(256 \right)} + \log{\left(40 \right)}^{2}}}{2}}}
=
1.52913018263282

y1=210e−log⁡(256)+log⁡(40)22y_{1} = 2 \sqrt{10} e^{\frac{\sqrt{- \log{\left(256 \right)} + \log{\left(40 \right)}^{2}}}{2}}
=
210e−log⁡(256)+log⁡(40)222 \sqrt{10} e^{\frac{\sqrt{- \log{\left(256 \right)} + \log{\left(40 \right)}^{2}}}{2}}
=
26.1586622606120
x2=210e−log⁡(256)+log⁡(40)22x_{2} = 2 \sqrt{10} e^{\frac{\sqrt{- \log{\left(256 \right)} + \log{\left(40 \right)}^{2}}}{2}}
=
210e−log⁡(256)+log⁡(40)222 \sqrt{10} e^{\frac{\sqrt{- \log{\left(256 \right)} + \log{\left(40 \right)}^{2}}}{2}}
=
26.1586622606120

y2=210e−log⁡(256)+log⁡(40)22y_{2} = \frac{2 \sqrt{10}}{e^{\frac{\sqrt{- \log{\left(256 \right)} + \log{\left(40 \right)}^{2}}}{2}}}
=
210e−log⁡(256)+log⁡(40)22\frac{2 \sqrt{10}}{e^{\frac{\sqrt{- \log{\left(256 \right)} + \log{\left(40 \right)}^{2}}}{2}}}
=
1.52913018263282
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