If \({x}^{y}={y}^{x}\) and \(x=2y\) determine the value for \(x \) and \(y\)
Answer:
we have \({x}^{y}={y}^{x} \) hence by using \(x=2y\) we get :
\({\left(2y\right)}^{y}={y}^{2y}={\left({y}^{2}\right)}^{y}\) thus \({\left(\frac{2y}{{y}^{2}}\right)}^{y}=1\)
hence \({\left(\frac{2}{y}\right)}^{y}=1\) take logarithm both sides to get :\(y\ln{\left(\frac{2}{y}\right)}=0\)
so \(y=0\) or \(\ln{\left(\frac{2}{y}\right)}=0=\ln{(1)}\) thus \(\frac{2}{y}=1\) i.e \(y=2\) and \(x=4\)
Answer:
we have \({x}^{y}={y}^{x} \) hence by using \(x=2y\) we get :
\({\left(2y\right)}^{y}={y}^{2y}={\left({y}^{2}\right)}^{y}\) thus \({\left(\frac{2y}{{y}^{2}}\right)}^{y}=1\)
hence \({\left(\frac{2}{y}\right)}^{y}=1\) take logarithm both sides to get :\(y\ln{\left(\frac{2}{y}\right)}=0\)
so \(y=0\) or \(\ln{\left(\frac{2}{y}\right)}=0=\ln{(1)}\) thus \(\frac{2}{y}=1\) i.e \(y=2\) and \(x=4\)
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