Question

Найти x
1. $(\frac{1}{3} )^{x - 1} \leqslant \frac{1}{ \sqrt[5]{5} }$
2. ${3}^{x^{2} + 2 } - {3}^{x^{2} - 1 } = {5}^{ {x}^{2} + 1} + {5}^{ {x}^{2} - 1}$
Помогите, пожалуйста

Answer 1

$1)\; \; (\frac{1}{3})^{x-1}\leq \frac{1}{\sqrt[5]5}\\\\3^{1-x}\leq 3^{log_3\frac{1}{\sqrt[5]5}}\\\\1-x\leq log_3\frac{1}{\sqrt[5]5}\; \; ,\; \; x\geq 1-log_3\frac{1}{\sqrt[5]5}\; ,\; \; x\geq 1+log_3\sqrt[5]5\; ,\\\\x\geq 1+\frac{1}{5}log_35$

$2)\; \; 3^{x^2+2}-3^{x^2-1}=5^{x^2+1}+5^{x^2-1}\\\\3^{x^2}\cdot (3^2-3^{-1})=5^{x^2}\cdot (5+5^{-1})\\\\3^{x^2}\cdot \frac{26}{3}=5^{x^2}\cdot \frac{26}{5}\\\\\frac{5^{x^2}}{3^{x^2}}=\frac{26/3}{26/5}\\\\(\frac{5}{3})^{x^2}=\frac{5}{3}\\\\x^2=1\\\\x^2-1=0\\\\(x-1)(x+1)=0\\\\x_1=-1\; ,\; \; x_2=1\\\\Otvet:\; \; x=-1\; ,\; x=1\; .$