Find algebraically the exact solutions to the equations (a) $ ext{ln}(4 - 2x) + ext{ln}(9 - 3x) = 2 ext{ln}(x + 1),$ $-1 < x < 2$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 7

To solve the equation, we will use the properties of logarithms. The left-hand side can be rewritten using the product rule of logarithms:

$ext{ln}((4 - 2x)(9 - 3x)) = 2 ext{ln}(x + 1)$

Using the power rule, we rewrite the equation as:

$ext{ln}((4 - 2x)(9 - 3x)) = ext{ln}((x + 1)^2)$

Next, we can remove the logarithms by exponentiating both sides:

$(4 - 2x)(9 - 3x) = (x + 1)^2$

Expanding both sides yields:

$36 - 30x - 6x + 6x^2 = x^2 + 2x + 1$

This simplifies to:

$6x^2 - 32x + 35 = 0$

We now apply the quadratic formula, where $a = 6$, $b = -32$, and $c = 35$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values:

$x = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 6 \cdot 35}}{2 \cdot 6}$
$x = \frac{32 \pm \sqrt{1024 - 840}}{12}$
$x = \frac{32 \pm \sqrt{184}}{12}$
$x = \frac{32 \pm 2\sqrt{46}}{12}$
$x = \frac{16 \pm \sqrt{46}}{6}$

Examining the valid range of solutions $-1 < x < 2$, we find that:

$x = \frac{16 + \sqrt{46}}{6} ext{ is valid, and } x = \frac{16 - \sqrt{46}}{6} ext{ is not valid.}$