Question

Решить данные уравнения:

$7^{(x^2-9)} =1$

$\frac{3^x}{\sqrt{3} } =\frac{1}{9}$

$6^{x+3} = \frac{1}{216}$

log(3)x=log(9)36

Answer 1

$7^{(x^2-9)} =1;$;

$7^{(x^2-9)} =7^0;$

x²-9=0;

x=3;x=-3;

**********

$\frac{3^x}{\sqrt{3} } =\frac{1}{9}$

$3^{x-0,5}=3^{-2}$

x-0,5=-2;

x=-1,5;

***********

$6^{x+3}=\frac{1}{216}$

$6^{x+3}  = \frac{1}{6^3}$

$6^{x+3}  = 6^{-3}$

x+3=-3;

x=-6;

*************

$log_3x=log_936=log_{3^2}6^2=\frac{2}{2}log_36=log_36$

x=6;