Question

1 Logaritmo da raiz cúbica de 128 na base 16
2 Logaritmo da raiz cúbica de 81 na base raiz de 27
(Estou dando 50 pts pq é urgente)

Answer 1

Resposta:

$log16(\sqrt[3]{128} )= log 16(\sqrt[3]{16.8})= \frac{log16(16.8)}{3} \\\frac{1}{3}[ log16(16) +log16(8)]\\\\log16(8)= x, 16^{x} = 8, 2^{4x} = 2^{3} , 4x= 3, x= \frac{3}{4} \\\\\\\\\frac{1}{3}[ 1 +\frac{3}{4} ]\\\\= \frac{1}{3} .\frac{7}{4} = 7/12$

$log 27 (\sqrt[3]{81} )= x\\\\27^{x} =\sqrt[3]{81}\\\\3^{3x} = 3^{\frac{4}{3} } \\\\3x= \frac{4}{3}\\\\x= 4/9$

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{log_{16}\:\sqrt[3]{128} = x}$

$\mathsf{16^x = 128^{\frac{1}{3}}}$

$\mathsf{(2^4)^x = (2^7)^{\frac{1}{3}}}$

$\mathsf{\not2^{4x} = \not2^{\frac{7}{3}}}$

$\mathsf{4x = \dfrac{7}{3}}$

$\boxed{\boxed{\mathsf{x = \dfrac{7}{12}}}}$

$\mathsf{log_{\sqrt{27}}\:\sqrt[3]{81} = x}$

$\mathsf{27^{\frac{x}{2}} = 81^{\frac{1}{3}}}$

$\mathsf{(3^3)^{\frac{x}{2}} = (3^4)^{\frac{1}{3}}}$

$\mathsf{\not3^{\frac{3x}{2}} = \not3^{\frac{4}{3}}}$

$\mathsf{\dfrac{3x}{2} = \dfrac{4}{3}}$

$\boxed{\boxed{\mathsf{x = \dfrac{8}{9}}}}$