(a) Simplify : \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\) (b) If \(123_{y} =
(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 \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\)
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) \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\)
\(\implies 9^{2x - 3} = 3^{x + 3}\)
\((3^{2})^{2x - 3} = 3^{x + 3}\)
\(3^{4x - 6} = 3^{x + 3}\)
\(\implies 4x - 6 = x + 3\)
\(4x - x = 3 + 6 \implies 3x = 9\)
\(x = 3\)