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

To solve the equation $2^y \times 4^{y+1} = 8$, we start by rewriting $4$ in terms of $2$:

$4 = 2^2$

Then, we rewrite the equation as:

$2^y \times (2^2)^{y+1} = 8$

This simplifies to:

$2^y \times 2^{2(y+1)} = 8$

Combining the exponents gives:

$2^{y + 2(y + 1)} = 8$

Simplifying the exponent further:

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

Equating the exponents leads to:

$3y + 2 = 3$

Solving for $y$, we get:

$3y = 3 - 2$ $3y = 1$ $y = \frac{1}{3}$