Question

Calcule o valor das expressões a seguir.

Answer 1

4)
a)
$log_{ \sqrt[3]{100} } \sqrt{0.1} = x$

$( \sqrt[3]{100})^{x} = \sqrt{0.1 } \\ ( \sqrt[3]{ {10}^{2} } )^{x} = \sqrt{ \frac{1}{10} }$

$( {10}^{ \frac{2}{3} } ) ^{x} = \sqrt{ {10}^{ - 1} } \\ {10}^{ \frac{2x}{3} } = {10}^{ - \frac{ 1}{2} } \\ \frac{2x}{3} = - \frac{1}{2} \\ 4x = - 3 \\ x = - \frac{3}{4}$

então

$log_{ \sqrt[3]{100} } \sqrt{0.1} = - \frac{3}{4}$

agora

$log_{ \sqrt[3]{0.5 } } \sqrt{8} = x \\ (\sqrt[3]{0.5}) ^{x} = \sqrt{8} \\ ( \sqrt[3]{ \frac{1}{2} } ) ^{x} = \sqrt{ {2}^{3} } \\ ( \sqrt[3]{ {2}^{ - 1} } ) ^{x} = {2}^{ \frac{3}{2} } \\ ( {2}^{ - \frac{1}{3} } ) ^{x} = {2}^{ \frac{3}{2} } \\ {2}^{ - \frac{x}{3} } = {2}^{ \frac{3}{2} } \\ - \frac{x}{3} = \frac{3}{2} \\ - 2x = 9 \\ x = - \frac{9}{2}$

então

$log_{ \sqrt[3]{0.5} }\sqrt{8} = - \frac{9}{2}$

(-3/4)+(-9/2)

$- \frac{3}{4} - \frac{9}{2} = \frac{ - 3 - 18}{4} = - \frac{21}{4}$

a) -21/4

b) faz do mesmo jeito da a)

$log_{0.1}0.01 = x \\ 0.1 ^{x} = 0.01$
de cara já sabemos de 0,1²=0,01

então

$log_{0.1}0.01 = 2$

$log_{ \sqrt{2} }0.25 = x \\ (\sqrt{2} ) ^{x} = 0.25 \\ ({2}^{ \frac{1}{2} } ) ^{x} = \frac{1}{4} \\ {2}^{ \frac{x}{2} } = {2}^{ - 2} \\ \frac{x}{2} = - 2 \\ x = - 2 \times 2 \\ x = - 4$

então

$3 log_{ \sqrt{2} }0.25 = - {4} \times 3 \\ 3 log_{ \sqrt{2} }0.25 = - 12$

$log_{25}0.008 = x \\ {25}^{x} = 0.008 \\ ( \frac{100}{4} ) ^{x} = \frac{8}{1000} \\ ( { \frac{10}{2} }^{2} )^{x} = { \frac{2}{10} }^{3} \\ \frac{10}{2} ^{2x} = \frac{10}{2} ^{ - 3} \\ 2x = - 3 \\ x = - \frac{3}{2}$

então

$log_{25}0.008 = - \frac{3}{2} \\ \frac{1}{2} log_{25}0.008 = - \frac{3}{2} \times \frac{1}{2} \\ \frac{1}{2} log_{25}0.008 = - \frac{3}{4}$

(2)-(-12)-(-3/4)

$2 + 12 + \frac{3}{4} = \frac{8 + 48 + 3}{4} = \frac{59}{4}$

b)59/4

5)
$log_{x}9 = - 2 \\ {x}^{ - 2} = 9 \\ \frac{1}{x} ^{2} = 9 \\ \frac{1}{ {x}^{2} } = 9 \\ 1 = 9 {x}^{2} \\ \frac{1}{9} = {x}^{2} \\ ( \frac{1}{3} ) ^{2} = {x}^{2} \\ x = \frac{1}{3}$

x=1/3