Question

2 ^ (x * log_2(7)) * 7 ^ (x ^ 2 + x) = 1​

Answer 1

Explicație pas cu pas:

$2^{xlog_2 \ 7} \cdot 7^{x ^ 2 + x} = 1$

$\boxed{nlog_a \ m = log_a \ m^{n}, \ a > 0, \ a \neq 1}$

$2^{log_ {2} \ 7^{x}} \cdot 7^{x ^ 2 + x} = 1$

$\boxed{a^{log_{a} \ b} = b, \ a > 0, \ a \neq 1}$

$7^{x} \cdot 7^{x ^ 2 + x} = 1$

$7^{x + x ^ 2 + x} = 7^{0}$

$x(x + 2) = 0$

$\implies x_{1} = -2; \ x_{2} = 0$