Question

Найти корень: 3^(log27 (7x+1)) =4

Answer 1

$27=3^3,; ; log_{3^3}{(7x+1)}=frac{1}{3}log_3{(7x+1)}=log_3{sqrt[3]{7x+1}}\\3^{log_3{sqrt[3]{7x+1}}}=4\\sqrt[3]{7x+1}=4\\7x+1=4^3\\7x=64-1\\x=9$