Question

Se log de 5 na base y (Logy5) = 2x?
0 < y ≠ 1, então:

(y^3x + y^-3x) / (y^x + y^-x) é igual a:

Answer 1

Se $\mathtt{log_y\,5=2x},$ então

$\mathtt{y^{2x}=5\quad\quad\quad(i)}$


Podemos elevar os dois lados a $\mathtt{3/2,}$ obtendo assim

$\mathtt{(y^{2x})^{3/2}=5^{3/2}}\\\\ \mathtt{y^{3x}=5^{3/2}\quad\quad\quad(ii)}$


Elevando os dois lados a $\mathtt{(-1),}$ temos

$\mathtt{y^{-3x}=5^{-3/2}\quad\quad\quad(iii)}$


Podemos elevar a $\mathtt{1/2}$ os dois lados da equação $\mathtt{(i)}$ obtendo assim

$\mathtt{(y^{2x})^{1/2}=5^{1/2}}\\\\ \mathtt{y^x=5^{1/2}\quad\quad\quad(iv)}$


Elevando os dois lados a $\mathtt{(-1),}$ obtemos também

$\mathtt{y^{-x}=5^{-1/2}\quad\quad\quad(v)}$

___________

Dessa forma, a expressão pedida é

$\mathtt{\dfrac{y^{3x}+y^{-3x}}{y^x+y^{-x}}}\\\\\\ =\mathtt{\dfrac{5^{3/2}+5^{-3/2}}{5^{1/2}+5^{-1/2}}}$


Multiplicando o numerador e o denominador por $\mathtt{5^{3/2}},$

$=\mathtt{\dfrac{(5^{3/2}+5^{-3/2})\cdot 5^{3/2}}{(5^{1/2}+5^{-1/2})\cdot 5^{3/2}}}\\\\\\ =\mathtt{\dfrac{5^{3/2}\cdot 5^{3/2}+5^{-3/2}\cdot 5^{3/2}}{5^{1/2}\cdot 5^{3/2}+5^{-1/2}\cdot 5^{3/2}}}\\\\\\ =\mathtt{\dfrac{5^{3/2+3/2}+5^{-3/2+3/2}}{5^{1/2+3/2}+5^{-1/2+3/2}}}\\\\\\ =\mathtt{\dfrac{5^3+5^0}{5^2+5^1}}\\\\\\ =\mathtt{\dfrac{125+1}{25+5}}\\\\\\ =\mathtt{\dfrac{126}{30}}\begin{array}{c}\mathtt{^{\div 6}}\\ \mathtt{^{\div 6}}\end{array}\\\\\\ =\boxed{\begin{array}{c}\mathtt{\dfrac{21}{5}} \end{array}}$


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Bons estudos! :-)