WAEC - Further Mathematics (2010 - No. 4)
If \(16^{3x} = \frac{1}{4}(32^{x - 1})\), find the value of x.
\(-1\)
\(\frac{-1}{3}\)
\(\frac{-3}{7}\)
\(\frac{-5}{19}\)
स्पष्टीकरण
\(16^{3x} = \frac{1}{4}(32^{x - 1})\)
\((2^{4})^{3x} = (2^{-2})((2^{5})^{x - 1})\)
\(2^{12x} = 2^{-2 + 5x - 5}\)
\(12x = -7 + 5x\)
\(7x = -7 \implies x = -1\)