7. (i) Use logarithms to solve the equation $8^{2x+1} = 24$, giving your answer to 3 decimal places - Edexcel - A-Level Maths Pure - Question 9 - 2015 - Paper 2

Start by rewriting the equation:

$\log_{2}(11y - 3) - \log_{2} 3 - 2 \log_{2} y = 1$

Using the properties of logarithms, combine the logarithmic terms:

$\log_{2}\left(\frac{11y - 3}{3 y^{2}}\right) = 1$

This implies:

$\frac{11y - 3}{3y^{2}} = 2$

Next, cross-multiply to eliminate the fraction:

$11y - 3 = 6y^{2}$

Reorganizing gives a quadratic equation:

$6y^{2} - 11y + 3 = 0$

Using the quadratic formula to solve for $y$:

$y = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{11 \pm \sqrt{(-11)^{2} - 4 \cdot 6 \cdot 3}}{2 \cdot 6}$

Calculating the discriminant:

$\sqrt{121 - 72} = \sqrt{49} = 7$

Now substituting back gives:

$y = \frac{11 \pm 7}{12}$

Thus, we have:

$y_{1} = \frac{18}{12} = 1.5$ $y_{2} = \frac{4}{12} = \frac{1}{3}$

Considering the constraint $y > \frac{3}{11}$, we find:

Only $y = 1.5$ is valid.

The final answer is:

$y = 1.5$