Question

B.143 IEZZI.

Se x=[tex3]10^{\frac{1}{1-logz}}[/tex3] e y=[tex3]10^{\frac{1}{1-logx}}[/tex3], prove que:
z=[tex3]10^{\frac{1}{1-logy}}[/tex3]

Answer 1

$\mathrm{x=10^{\frac{1}{1-\log{z}}}\ \bigg|\ y=10^{\frac{1}{1-\log{x}}}}\\\\\\ \mathrm{\log{x}=\log{10^{\frac{1}{1-log{z}}}}\ \to\ \log{x}=\bigg(\dfrac{1}{1-\log{z}}\bigg)\log{10}}\\\\ \mathrm{\log{x}=\dfrac{1}{1-\log{z}}}\\\\\\ \mathrm{\log{y}=\log{10^{\frac{1}{1-\log{x}}}}\ \to\ \log{y}=\bigg(\dfrac{1}{1-\log{x}}\bigg)\log{10}}\\\\ \mathrm{\log{y}=\dfrac{1}{1-\log{x}}\ \to\ 1-\log{x}=\dfrac{1}{\log{y}}\ \to\ \log{x}=1-\dfrac{1}{\log{y}}}\\\\ \mathrm{\log{x}=\dfrac{\log{y}-1}{\log{y}}}$

$\mathrm{\dfrac{1}{1-\log{z}}=\dfrac{\log{y}-1}{\log{y}}\ \to\ \dfrac{\log{y}}{\log{y}-1}=1-\log{z}}\\\\ \mathrm{\log{z}=1-\dfrac{\log{y}}{\log{y}-1}\ \to\ \log{z}=\dfrac{\log{y}-1-\log{y}}{\log{y}-1}}\\\\ \mathrm{\log{z}=-\dfrac{1}{\log{y}-1}\ \to\ \log{z}=\dfrac{1}{1-\log{y}}\ \to\ \boxed{\mathbf{z=10^{\frac{1}{1-\log{y}}}}}}$