Question

ПОЖАЛУЙСТА РЕШИТЕ УРОВНЕНИЯ А) 2^x=32 Б) log1/3×(4x+5)=-1 В) 3^x+3^x-1=4

Answer 1

$1) 2^x = 32 \\ 2^x = 2^5 \\ x= 5 \\ 2) log_{ \frac{1}{3}}(4x+5) = -1 \\ log_3(4x+5) = 1 \\ log_3(4x+5) = log_33 \\ 4x + 5 = 3 \\ 4x = -2 \\ x = - \frac{1}{2} \\ 3) 3^x + 3^{x-1} = 4\\ 3^x + 3^{x-1} = 3^{log_34} \\ x + x - 1 = log_34 \\ 2x = log_34 + log_33 \\ 2x = log_312 \\ x = \frac{log_312}{2} = \frac{log_312}{log_39} = log_912 \\ c:$