WAEC - Further Mathematics (2021 - No. 5)
Solve (\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\)
\(\frac{7}{5}\)
\(\frac{6}{7}\)
\(\frac{-7}{6}\)
\(\frac{-6}{7}\)
Selitys
(\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\)
3\(^{-2(x + 2)}\) = 3\(^{5(x - 2)}\);
-2(x+2) = 5(x - 2)
-2x -4 = 5x - 10
-7x = -6
x = \(\frac{6}{7}\)