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 6 - 2012 - Paper 4

Starting from the equation:
$1 + 2 ext{log}_3 x = ext{log}_3 (28x - 9)$

1. Replace $1$ with $ext{log}_3 3$:
   $ext{log}_3 3 + 2 ext{log}_3 x = ext{log}_3 (28x - 9)$

2. Use the property of logarithms to combine the left side:
   $ext{log}_3 (3x^2) = ext{log}_3 (28x - 9)$

3. Exponentiating both sides to eliminate the logarithm gives:
   $3x^2 = 28x - 9$

4. Rearranging leads to:
   $3x^2 - 28x + 9 = 0$

5. Now we can use the quadratic formula $x = \frac{-b \\pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 3, b = -28, c = 9$:

   • Calculate $b^2 - 4ac$:
     $(-28)^2 - 4(3)(9) = 784 - 108 = 676$
   • Compute the two solutions:
     $x = \frac{28 \\pm \sqrt{676}}{6}$
   • Simplifying gives:
     $x = \frac{28 \\pm 26}{6}$
   • Thus the two potential solutions are:
     $x = \frac{54}{6} = 9$
     $x = \frac{2}{6} = \frac{1}{3}$