Question

Log2 na base 0,25 + log8 na base16

Answer 1

$\mathrm{\ell og_{\,0,25}}2+\mathrm{\ell og_{16}}8\\ \\ =\mathrm{\ell og_{1/4}}2+\mathrm{\ell og_{16}}8\\ \\ =x+y$

onde

$x=\mathrm{\ell og_{1/4}}2\\ \\ y=\mathrm{\ell og_{16}}8$.


$x=\mathrm{\ell og_{1/4}}2\\ \\ \left(\dfrac{1}{4} \right )^{x}=2\\ \\ \left(4^{-1} \right )^{x}=2\\ \\ 4^{-x}=2\\ \\ \left(2^{2} \right )^{-x}=2\\ \\ 2^{-2x}=2^{1}\\ \\ -2x=1\\ \\ x=-\dfrac{1}{2} \Rightarrow \boxed{\mathrm{\ell og_{1/4}}2=-\dfrac{1}{2}}$


$y=\mathrm{\ell og_{16}}8\\ \\ 16^{y}=8\\ \\ \left(2^{4} \right )^{y}=2^{3}\\ \\ 2^{4y}=2^{3}\\ \\ 4y=3\\ \\ y=\dfrac{3}{4} \Rightarrow \boxed{\mathrm{\ell og_{16}}8=\dfrac{3}{4}}$



Então

$x+y\\ \\ =-\dfrac{1}{2}+\dfrac{3}{4}\\ \\ =-\dfrac{2}{4}+\dfrac{3}{4}\\ \\ =\dfrac{-2+3}{4}\\ \\ =\dfrac{1}{4}\\ \\ =0,25\\ \\ \\ \boxed{\mathrm{\ell og_{\,0,25}}2+\mathrm{\ell og_{16}}8=0,25}$

Answer 2

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\mathsf{log_{0,25}\:2 + log_{16}\:8 = log_{\frac{1}{4}}\:2 + log_{2^4}\:2^3}$

$\mathsf{log_{0,25}\:2 + log_{16}\:8 = log_{2^{-2}}\:2 + log_{2^4}\:2^3}$

$\mathsf{log_{0,25}\:2 + log_{16}\:8 = -\dfrac{1}{2}\:.\:log_{2}\:2 + \dfrac{3}{4}\:.\:log_{2}\:2}$

$\mathsf{log_{0,25}\:2 + log_{16}\:8 = \dfrac{3}{4} - \dfrac{1}{2}}$

$\boxed{\boxed{\mathsf{log_{0,25}\:2 + log_{16}\:8 = \dfrac{1}{4}}}}$