If 2$^{x + y}$ = 16 and 4$^{x - y} = \frac{1}{32}$, find the values of x and y.

To solve this problem, we need to express both sides of the equations with the same base so that we can equate the exponents.

1. Analyze the equations

We are given: (i) $2^{x + y} = 16$ (ii) $4^{x - y} = \frac{1}{32}$

2. Simplify Equation (i)

Since $16 = 2^4$, we can rewrite the equation as: $2^{x + y} = 2^4$ Equating the exponents: $x + y = 4$ --- (Equation 1)

3. Simplify Equation (ii)

We know that $4 = 2^2$ and $32 = 2^5$. Substituting these into the equation: $(2^2)^{x - y} = \frac{1}{2^5}$ $2^{2(x - y)} = 2^{-5}$ Equating the exponents: $2(x - y) = -5$ $x - y = -\frac{5}{2}$ $x - y = -2.5$ --- (Equation 2)

4. Solve the simultaneous equations

We now have:

1. $x + y = 4$
2. $x - y = -2.5$

Add Equation 1 and Equation 2: $(x + y) + (x - y) = 4 + (-2.5)$ $2x = 1.5$ $x = \frac{1.5}{2} = 0.75$ or $\frac{3}{4}$

Now, substitute $x = \frac{3}{4}$ into Equation 1: $\frac{3}{4} + y = 4$ $y = 4 - \frac{3}{4}$ $y = \frac{16 - 3}{4}$ $y = \frac{13}{4}$

5. Conclusion

The values are $x = \frac{3}{4}$ and $y = \frac{13}{4}$. This matches option B.

Exam Tip: When dealing with indices, always try to reduce the numbers to their smallest prime base (like 2, 3, or 5). Remember that $\frac{1}{a^n} = a^{-n}$; this is a common trap in JAMB questions that tests your knowledge of negative indices.