(a) Simplify : \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\)
(b) If \(123_{y} = 83_{10}\), obtain an equation in y, hence find the value of y.
(c) Solve the equation \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\)
Show Answer Show Explanation Explanation (a) \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\) = \(\log_{10} (25^{\frac{1}{2}}) - \log_{10} (3^{2}) + \log_{10} 18\) = \(\log_{10} (\frac{5 \times 18}{9})\) = \(\log_{10} 10\) = 1 (b) \(123_{y} = (1 \times y^{2}) + (2 \times y^{1}) + (3 \times y^{0})\) \(y^{2} + 2y + 3 = 83\) \(y^{2} + 2y + 3 - 83 = 0 \implies y^{2} + 2y - 80 = 0\) Equation : \(y^{2} + 2y - 80 = 0\) \(y^{2} - 8y + 10y - 80 = 0\) \(y(y - 8) + 10(y - 8) = 0\) \((y + 10)(y - 8) = 0 \implies y = -10 ; 8\) y = 8 since y cannot be negative. (c) \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\) \(\implies 9^{2x - 3} = 3^{x + 3}\) \((3^{2})^{2x - 3} = 3^{x + 3}\) \(3^{4x - 6} = 3^{x + 3}\) \(\implies 4x - 6 = x + 3\) \(4x - x = 3 + 6 \implies 3x = 9\) \(x = 3\)