Question

Va rog ajutați-ma ch rezolvarea logaritmilor

Answer 1

$log_3 (6-x) + 2log_3 (\sqrt[]{4-x}) =1$, $D = (- \infty; 4)$

$2log_3 \sqrt{4-x} = log_3 ( \sqrt{4-x} )^2 = log_3 (4-x)$
$log_3 (6-x) + log_3 (4-x) = log_3 [(6-x)(4-x)]$
1 = $log_3 3$

$log_3 [(6-x)(4-x)] = log_3 3 =\ \textgreater \ (6-x)(4-x) = 3$
$x^2 -10x +24 = 3 =\ \textgreater \ x^2 - 10x + 21 = 0$
(x-3)(x-7) = 0 => x1 = 3 ∈ D ; x2 = 7 ∉ D