Question

(MACK-SP) Se
$log \sqrt{0.1} = x \:$
então
${x}^{2}$
é
​

Answer 1

Explicação passo-a-passo:

Pela definição:

${10}^{x} = \sqrt{0.1}$

${10}^{x} = \sqrt{ \frac{1}{10} }$

${10}^{x} = \frac{ \sqrt{1} }{ \sqrt{10} }$

${10}^{x} = \frac{1}{ {10}^{ \frac{1}{2} } }$

${10}^{x} = {10}^{ - \frac{1}{2} }$

$x = - \frac{1}{2}$

${x}^{2} = ( - \frac{1}{2} {)}^{2}$

${x}^{2} = \frac{1}{4}$

Answer 2

$log( \sqrt{0,1} ) = log( {0,1})^{ \frac{1}{2} } \\ \frac{1}{2} log(0,1) = \frac{1}{2}. log( \frac{1}{10} ) \\ \frac{1}{2} ( log(1) - log(10) )$

$\frac{1}{2}(0 - 1) = - \frac{1}{2}$

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