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 6 - 2010 - Paper 4

First, we rearrange the equation. Using the properties of logarithms, we have:

[ log_2(11 - 6x) = log_2((x - 1)^2) + 3 ]

This can be rewritten as: [ log_2(11 - 6x) = log_2((x - 1)^2) + log_2(2^3) ]

Combining the logs gives: [ log_2(11 - 6x) = log_2((x - 1)^2 \cdot 8) ]

Setting the arguments equal to each other: [ 11 - 6x = (x - 1)^2 \cdot 8 ]

Expanding the right-hand side: [ 11 - 6x = 8(x^2 - 2x + 1) ]

Which expands to: [ 11 - 6x = 8x^2 - 16x + 8 ]

Rearranging yields: [ 0 = 8x^2 - 10x - 3 ]

Finding the roots using the quadratic formula: [ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

Where a = 8, b = -10, and c = -3: [ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} ] [ = \frac{10 \pm \sqrt{100 + 96}}{16} ] [ = \frac{10 \pm \sqrt{196}}{16} ] [ = \frac{10 \pm 14}{16} ]

Calculating the two possible values for x:

1. [ x = \frac{24}{16} = \frac{3}{2} ] (valid)
2. [ x = \frac{-4}{16} = -\frac{1}{4} ] (not valid as x must be positive)

Thus, the solution for x is: [ x = \frac{3}{2} ]