Given that $y = 3x^2$, (a) show that $ ext{log}_3 y = 1 + 2 ext{log}_3 x$ (b) Hence, or otherwise, solve the equation $1 + 2 ext{log}_3 x = ext{log}_3 (28x - 9)$ - Edexcel - A-Level Maths Pure - Question 7 - 2012 - Paper 4

To solve the equation, start with: $1 + 2 ext{log}_3 x = ext{log}_3 (28x - 9)$ Substituting $ext{log}_3 3$ into the left side gives: $ext{log}_3 3 + 2 ext{log}_3 x = ext{log}_3 (28x - 9)$ This can be combined as: $ext{log}_3 (3x^2) = ext{log}_3 (28x - 9)$ Now we remove the logarithms (since the bases are equal) to compare the arguments: $3x^2 = 28x - 9$ Rearranging this gives: $3x^2 - 28x + 9 = 0$ Applying the quadratic formula: $x = \frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$ Here, $a = 3$, $b = -28$, and $c = 9$: $x = \frac{28 \\pm \\sqrt{(-28)^2 - 4 * 3 * 9}}{2 * 3}$ Calculating the discriminant: $(-28)^2 - 4 * 3 * 9 = 784 - 108 = 676$, so: $x = \frac{28 \\pm 26}{6}$ Calculating the two potential solutions:

1. $x = \frac{54}{6} = 9$
2. $x = \frac{2}{6} = \frac{1}{3}$ Thus, the solutions to the equation are: $x = 9 \\text{ or } \\ x = \frac{1}{3}$