(a) Simplify : (frac{1}{2}log_{10} 25 – 2log_{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 – 2log_{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)