Question

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

Answer 1

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