Question

3^x+2/4=5^x+2
2^x^2-6x+0.5=1/16√2- показникові рівняння

Answer 1

$1)3^{\frac{x+2}{4}}=5^{x+2} \\\\(3^{\frac{1}{4}})^{x+2}=5^{x+2}|:5^{x+2}>0\\\\(\frac{\sqrt[4]{3}}{5})^{x+2}=1\\\\(\frac{\sqrt[4]{3}}{5})^{x+2}=(\frac{\sqrt[4]{3}}{5})^{o}\\\\x+2=0\\\\x=-2\\\\Otvet:{-2}$

$2)2^{x^{2}-6x+0,5}=\frac{1}{16\sqrt{2}}\\\\2^{x^{2}-6x+0,5} =\frac{1}{2^{4}*2^{\frac{1}{2}}}\\\\2^{x^{2}-6x+0,5}=2^{-4,5}\\\\x^{2}-6x+0,5=-4,5\\\\x^{2}-6x+5=0\\\\x_{1}=1\\\\x_{2}=5\\\\Otvet:\boxed{1;5}$