Find algebraically the exact solutions to the equations (a) ln(4 - 2x) + ln(9 - 3x) = 2ln(x + 1), -1 < x < 2 - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 7

To solve the equation, we begin by applying the logarithm properties.

Using the property ( \ln a + \ln b = \ln(ab) ), we can combine the left-hand side:

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

Next, using the power rule of logarithms, we rewrite the right-hand side:

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

Now, we eliminate the logarithm by exponentiating both sides:

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

Expanding both sides gives:

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

Rearranging this results in:

$5x^2 - 32x + 35 = 0$

We can now use the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) with ( a = 5, b = -32, c = 35 ):

$x = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 5 \cdot 35}}{2 \cdot 5}$

Calculating the discriminant gives:

$x = \frac{32 \pm \sqrt{1024 - 700}}{10}$

$x = \frac{32 \pm \sqrt{324}}{10}$

Which simplifies to:

$x = \frac{32 \pm 18}{10}$

The two potential solutions are:

$x = 5 \quad \text{and} \quad x = 1.4$

Since the question states that (-1 < x < 2), we take ( x = 1.4 ) as our acceptable solution.