Question

3^x +16 = 19^6 : ( 19^3 . 19^2 ) - 3 . 1^2025 + 2022^0

 

Answer 1

\(3^x+16=19^6:\left(19^3.19^2\right)-3.1^{2025}+2022^0\)
\(\Rightarrow3^x+16=19^6:19^5-3+1\)
\(\Rightarrow3^x+16=19-2\)
\(\Rightarrow3^x+16=17\)
\(\Rightarrow3^x=1=3^0\)
Vậy \(x=0\)