Question

$\mathsf{Se~~x=10^{\dfrac{1}{1-log~z}}~~e~~y=10^{\dfrac{1}{1-log~x}},~~prove~que~~z=10^{\dfrac{1}{1-log~y}}.}$

Answer 1

$\large\begin{array}{l} \textsf{Provar o seguinte teorema:}\\\\ \textsf{Se }\mathsf{x=10^\frac{1}{1-\ell og\,z}}\textsf{ e }\mathsf{y=10^{\frac{1}{1-\ell og\,x}},}\textsf{ ent\~ao }\mathsf{z=10^{\frac{1}{1-\ell og\,y}}.}\\\\\\ \bullet~~\textsf{Condi\c{c}\~oes de exist\^encia, determinados pelas hip\'oteses:}\\\\ \mathsf{\ell og\,z\ne 1~\Rightarrow~z>0~~e~~z\ne 10\qquad(i)}\\\\ \mathsf{\ell og\,x\ne 1~\Rightarrow~x>0~~e~~x\ne 10\qquad(ii)} \end{array}$

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$\large\begin{array}{l} \textsf{Prosseguindo com a prova e usando as condi\c{c}\~oes (i) e (ii)}\\\textsf{sempre que necess\'arias:}\\\\ \textsf{Se }\mathsf{x=10^{\frac{1}{1-\ell og\,z}},}\textsf{ ent\~ao}\\\\ \mathsf{\ell og\,x=\ell og\!\left(10^{\frac{1}{1-\ell og\,z}}\right )}\\\\ \mathsf{\ell og\,x=\dfrac{1}{1-\ell og\,z}}\\\\ \mathsf{-\ell og\,x=\dfrac{-1}{1-\ell og\,z}} \end{array}$

$\large\begin{array}{l} \mathsf{1-\ell og\,x=1+\dfrac{-1}{1-\ell og\,z}}\\\\ \mathsf{1-\ell og\,x=\dfrac{1-\ell og\,z}{1-\ell og\,z}+\dfrac{-1}{1-\ell og\,z}}\\\\ \mathsf{1-\ell og\,x=\dfrac{1-\ell og\,z-1}{1-\ell og\,z}}\\\\ \mathsf{1-\ell og\,x=\dfrac{-\ell og\,z}{1-\ell og\,z}}\qquad\textsf{(usando (i) a seguir)} \end{array}$


$\large\begin{array}{l} \textsf{Como }\mathsf{\ell og\,x\ne 1,}\textsf{ os dois membros da igualdade acima nunca}\\\textsf{se anulam. Portanto, podemos obter os inversos:}\\\\ \mathsf{\dfrac{1}{1-\ell og\,x}=\dfrac{1}{~\frac{-\ell og\,z}{1-\ell og\,z}~}}\\\\ \mathsf{\dfrac{1}{1-\ell og\,x}=\dfrac{1-\ell og\,z}{-\ell og\,z}}\\\\ \mathsf{\dfrac{1}{1-\ell og\,x}=\dfrac{\ell og\,z-1}{\ell og\,z}} \end{array}$


$\large\begin{array}{l} \textsf{Da igualdade acima, obtemos como consequ\^encia mais uma}\\\textsf{condi\c{c}\~ao para z:}\\\\ \mathsf{\ell og\,z\ne 0~~\Rightarrow~~z\ne 1\qquad(iii)}\\\\\\ \textsf{Aplicando exponencial de base 10 aos dois membros, temos}\\\\ \mathsf{\dfrac{1}{1-\ell og\,x}=\dfrac{\ell og\,z-1}{\ell og\,z}}\\\\ \mathsf{10^{\frac{1}{1-\ell og\,x}}=10^{\frac{1-\ell og\,z}{-\ell og\,z}}} \end{array}$


$\large\begin{array}{l} \textsf{Mas por hip\'otese, }\mathsf{10^{\frac{1}{1-\ell og\,x}}=y.}\textsf{ Dessa forma, ficamos com}\\\\ \mathsf{y=10^{\frac{1-\ell og\,z}{-\ell og\,z}}}\\\\\\\textsf{e segue que}\\\\ \mathsf{\ell og\,y=\ell og\!\left(10^{\frac{1-\ell og\,z}{-\ell og\,z}}\right)}\\\\ \mathsf{\ell og\,y=\dfrac{1-\ell og\,z}{-\ell og\,z}}\\\\ \end{array}$

$\large\begin{array}{l} \mathsf{-\ell og\,y=-\,\dfrac{1-\ell og\,z}{-\ell og\,z}}\\\\ \mathsf{-\ell og\,y=\dfrac{1-\ell og\,z}{\ell og\,z}}\\\\ \mathsf{1-\ell og\,y=1+\dfrac{1-\ell og\,z}{\ell og\,z}}\\\\ \mathsf{1-\ell og\,y=\dfrac{\ell og\,z}{\ell og\,z}+\dfrac{1-\ell og\,z}{\ell og\,z}}\\\\ \mathsf{1-\ell og\,y=\dfrac{\ell og\,z+1-\ell og\,z}{\ell og\,z}}\\\\ \mathsf{1-\ell og\,y=\dfrac{1}{\ell og\,z}}\qquad\textsf{(usando (iii) a seguir)} \end{array}$


$\large\begin{array}{l} \\\\ \textsf{Observe que o lado direito da igualdade acima \'e o inverso}\\\textsf{de um n\'umero diferente de zero, e portanto ambos os}\\\textsf{membros n\~ao s\~ao nulos e possuem inversos.} \\\\\textsf{Tomando os inversos dos dois membros, ficamos com}\\\\ \mathsf{\dfrac{1}{1-\ell og\,y}=\ell og\,z}\\\\ \mathsf{10^{\frac{1}{1-\ell og\,y}}=10^{\ell og\,z}}\\\\ \mathsf{10^{\frac{1}{1-\ell og\,y}}=z}\\\\ \boxed{\begin{array}{c}\mathsf{z=10^{\frac{1}{1-\ell og\,y}}} \end{array}}\\\\\\ \textsf{como quer\'iamos demonstrar.} \end{array}$


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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: logaritmo log prova demonstração teorema exponencial base condição hipótese