Question

alguem responde por favor

Answer 1

$a)\\\\\log{x}+\log{y}=2\longrightarrow\log(x\cdot{y})=2\Longleftrightarrow10^2=x\cdot{y}\longrightarrow\boxed{\boxed{{x}y=100}}$ $b)\\\\\log_3x-\log_3y=2\longrightarrow\log_3\left(\dfrac{x}{y}\right)=2\Longleftrightarrow3^2=\dfrac{x}{y}\longrightarrow\boxed{\boxed{\dfrac{x}{y}=9}}$ Por comodidade, para respondermos a alternativa c), vamos determinar os valores de 'x' e 'y': $\begin{Bmatrix}\alpha)&xy=100&\\\\\beta)&\dfrac{x}{y}=9&\longrightarrow{x}=9y\end{matrix}\\\\\\\alpha)\ xy=100\longrightarrow(9y)y=100\longrightarrow9y^2=100\longrightarrow{y}^2=\dfrac{100}{9}\longrightarrow\\\\\\y=\sqrt{\dfrac{100}{9}}\left\langle\begin{matrix}\boxed{y=+\dfrac{10}{3}}\\\\\\y=-\dfrac{10}{3}\end{matrix}\right\\\\\\-\dfrac{10}{3}\text{ n}\mathrm{\tilde{a}o\ satisfaz\ a\ condi}\text{\c{c}}\mathrm{\tilde{a}o\ de\ exist\hat{e}ncia\ de\ }\log{y}\\\\\\\alpha)\ xy=100\\\\x\left(\dfrac{10}{3}\right)=100\\\\\\x=\dfrac{3\cdot100}{10}\\\\\\\boxed{x=30}\\\\\\\\\log(x-6y)=\ ?\\\\\\\log\left((30)-6\left(\dfrac{10}{3}\right)\right)=\\\\\\\log(30-2\cdot10)=\\\\\log(30-20)=\\\\\log(10)=\boxed{\boxed{1}}$