Question

Объясните пожалуйста подробно
log2(2x-x^2)=0
log16x=2
log1/3(x+2)=1

Answer 1

$log_{2}(2x - {x}^{2} ) = 0 \\ 2x - {x}^{2} = {2}^{0} \\ 2x - {x}^{2} = 1 \\ {x }^{2} - 2x + 1 = 0 \\ {(x - 1)}^{2} = 0 \\ x = 1$
$log_{16}(x) = 2 \\ {16}^{2} = x \\ x = 256$
$log_{ \frac{1}{3} }(x + 2) = 1 \\ { (\frac{1}{3} )}^{1} = x + 2 \\ x = - \frac{5}{3}$