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 simplify the right side:

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

Using the property of logarithms that states ( \log_b a + c = \log_b(a \cdot b^c) ), we can rewrite this as:

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

This leads to:

[ 11 - 6x = (x - 1)^2 \cdot 8 ]

Expanding the right side, we have:

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

This simplifies to:

[ 11 - 6x = 8x^2 - 16x + 8 ]

Rearranging gives:

[ 0 = 8x^2 - 10x - 3 ]

Now we can solve this quadratic equation using the quadratic formula:

[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} ]

Calculating the discriminant:

[ (-10)^2 - 4 \cdot 8 \cdot (-3) = 100 + 96 = 196 ]

Using this in the formula:

[ x = \frac{10 \pm 14}{16} ]

Thus, we find two values:

[ x = \frac{24}{16} = 1.5 \quad \text{and} \quad x = \frac{-4}{16} = -0.25 ]

Since we are looking for positive $x$, we take ( x = 1.5 ).