Question

Calcule o valor do logaritmo

Obs: a base é 1/4

Answer 1

Lembre-se:

$1/x^{n} = x^{-n}$
$\sqrt[n]{x^{y}} = x^{y/n}$
$log_{(x^{n})}(a) = (1 / n) * log_{x} (a)$
$log_{(x)} (x) = 1$
$log_{(x)}(a) = n <=> x^{n} = a$
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$log_{(1/4)} (2\sqrt{2}) =log_{(1/2^{2})} (2 \sqrt{2})$
$log_{(1/4)}(2 \sqrt{2})= log_{(2^{-2})}(2*2^{1/2})$
$log_{(1/4)}(2 \sqrt{2})= (1 / [-2]) * log_{(2)} (2^{1}*2^{1/2})$
$log_{(1/4)}(2 \sqrt{2})= - (1 / 2) * log_{(2)} (2^{1+1/2})$
$log_{(1/4)}(2 \sqrt{2})= - (1 / 2) * log_{(2)} (2^{3/2})$
$log_{(1/4)}(2 \sqrt{2})= - (1 / 2) * (3 / 2) * log_{(2)} 2$
$log_{(1/4)}(2 \sqrt{2})=- (3 / 4) * 1$
$log_{(1/4)}(2 \sqrt{2})=-3/4$
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Outro método:

$log_{(1/4)} (2 \sqrt{2} ) = x$
$(1/4)^{x} = 2 \sqrt{2}$
$(1/2^{2})^{x} = \sqrt{2*2^{2}}$
$(2^{-2})^{x} = \sqrt{2^{3}}$
$2^{-2x} = 2^{3/2}$

Bases iguais, iguale os expoentes:

$-2x=3/2$
$2x=-3/2$
$2*2x=-3$
$4x=-3$
$x=-3/4$

Answer 2

1/4 x= 2raiz2
(1/4)^2 x= (raiz 8)^2
1/16 x= 8
1/2^4 x= 2^3
2^(-4) x= 2^3
- 4x = 3 .(-1)
4x = - 3
x = -3/4

logo:

log1/4 na base por (2raiz2) = - 3 / 4