Solve (a) 2^y = 8 (b) 2^y × 4^{y+1} = 8 - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 2

First, we will express 4 in terms of powers of 2. We know that $4 = 2^2$, so we can rewrite the equation as:

$2^y × (2^2)^{y+1} = 8$

This simplifies to:

$2^y × 2^{2y + 2} = 8$

Combining the powers of 2 gives us:

$2^{y + 2y + 2} = 8$

Thus, we can simplify further:

$2^{3y + 2} = 8$

Since $8 = 2^3$, we have:

$2^{3y + 2} = 2^3$

This implies the exponents are equal:

$3y + 2 = 3$

Now, we can solve for $y$:

$3y = 3 - 2$ $3y = 1$ y = rac{1}{3}