Question

1)(x-2)^3=-216

2)log4(4+7x)=log4(1+5x)+1 Решите плиз

Answer 1

$1)(x-2)^3=-216 \ \ sqrt[3]{(x-2)^3} = sqrt[3]{-216} \ \ (x-2) = -6 \ \ x = -4$

$2) log_4(4+7x)=log_4(1+5x)+1 \ \ log_4(4+7x)=log_4(1+5x)+log_44 \ \ log_4(4+7x)=log_44(1+5x) \ \ 4+7x = 4(1+5x) \ \ 4+7x = 4+20x \ \ 4-4 = 20x - 7x \ \ 13x = 0 \ \ x = 0$

ОДЗ: 

$left { {{4+7x textgreater 0} atop {1+5x textgreater 0}} right. left { {{x textgreater - frac{4}{7} } atop {x textgreater -frac{1}{5} }} right. Rightarrow x textgreater -frac{1}{5}$

$x = 0 in ODZ$

Ответ: $x = 0$