Question

Alguém me ajuda?? Pfv

Se log₃ 729 = x e log₃ ∛81 = y . O valor de x+y é: *

a) 22/3
b) 11/3
c) 10/3
d) 8/3

Answer 1

$\large\boxed{\begin{array}{l}\sf \ell og_3729=x\\\sf x=\ell og_33^6=6\ell og_33=6\\\sf y=\ell og_3\sqrt[\sf3]{\sf81}=\ell og_33^{\frac{4}{3}}=\dfrac{4}{3}\ell og_33=\dfrac{4}{3}\\\\\sf x+y=6+\dfrac{4}{3}=\dfrac{18+4}{3}=\dfrac{22}{3}\\\\\huge\boxed{\boxed{\boxed{\boxed{\sf\red{\maltese}~\blue{alternativa~a}}}}}\end{array}}$

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{log_3\:729 = x}$

$\mathsf{3^x = 729}$

$\mathsf{\not3^x = \not3^6}$

$\mathsf{x = 6}$

$\mathsf{log_3\:\sqrt[3]{81} = y}$

$\mathsf{3^y = \sqrt[3]{3^4}}$

$\mathsf{\not3^y = \not3^{\frac{4}{3}}}$

$\mathsf{y = \dfrac{4}{3}}$

$\mathsf{x + y = 6 + \dfrac{4}{3}}$

$\boxed{\boxed{\mathsf{x + y = \dfrac{22}{3}}}}$