5. (a) Find the positive value of x such that $$\log_x 64 = 2$$ (b) Solve for x $$\log_2(11 - 6x) = 2 \log_2(x - 1) + 3$$ - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 4

First, we simplify the equation by using properties of logarithms:

$\log_2(11 - 6x) = \log_2((x - 1)^2) + \log_2(8)$

This gives:

$\log_2(11 - 6x) = \log_2\left((x - 1)^2 \cdot 8\right)$

Since the logs are equal, we set the arguments equal to each other:

$11 - 6x = 8(x - 1)^2$

Expanding the right side:

$11 - 6x = 8(x^2 - 2x + 1)$

This simplifies to:

$11 - 6x = 8x^2 - 16x + 8$

Rearranging the equation gives:

$0 = 8x^2 - 10x - 3$

We can now use the quadratic formula, where a = 8, b = -10, and c = -3:

$x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8}$

Calculating the discriminant:

$x = \frac{10 \pm \sqrt{100 + 96}}{16} \Rightarrow x = \frac{10 \pm \sqrt{196}}{16} \Rightarrow x = \frac{10 \pm 14}{16}$

This gives two solutions:

1. $x = \frac{24}{16} = \frac{3}{2}$
2. $x = \frac{-4}{16} = -\frac{1}{4}$

Since we are looking for positive solutions, we have:

$x = \frac{3}{2}$