If $4^{x + y} = 1$ and $4^{x - y} = 4$, then the values of $x$ and $y$ respectively are

### Concept & Logic Any non-zero base raised to the power of $0$ equals $1$ (i.e., $a^0 = 1$). Additionally, any base raised to the power of $1$ equals itself (i.e., $a^1 = a$). By equating the right side of the equations to powers of $4$, we can solve for the variables in the exponents. ### Step-by-Step Solution * **Step 1:** Translate the right side of the first equation into a power of $4$. Since $4^0 = 1$, we can rewrite the first equation: $$ 4^{x + y} = 4^0 $$ * **Step 2:** Translate the right side of the second equation into a power of $4$. Since $4^1 = 4$, we can rewrite the second equation: $$ 4^{x - y} = 4^1 $$ * **Step 3:** Equate the exponents to form a linear system of equations. (1) $x + y = 0$ (2) $x - y = 1$ * **Step 4:** Solve for $x$ by adding the two equations together. $(x + y) + (x - y) = 0 + 1$ $2x = 1$ $x = \frac{1}{2}$ * **Step 5:** Solve for $y$ by substituting $x$ back into equation (1). $\frac{1}{2} + y = 0$ $y = -\frac{1}{2}$ ### Exam Strategy & Shortcut **Option Verification:** Instead of solving the equations, plug the given options directly into the simpler of the two equations: $x + y = 0$. The only way two numbers add up to zero is if they are additive inverses (same number, opposite signs). Looking at the options: (a) $-1/2 + 1/2 = 0$ (Possible) (b) $-1/2 - 1/2 = -1 \neq 0$ (Eliminate) (c) $1/2 - 1/2 = 0$ (Possible) (d) $1/2 + 1/2 = 1 \neq 0$ (Eliminate) Now check the remaining options (a) and (c) in the second equation: $x - y = 1$. Test (a): $-1/2 - (1/2) = -1 \neq 1$. Test (c): $1/2 - (-1/2) = 1/2 + 1/2 = 1$. This is the correct pair. ### Common Pitfall A very common mistake happens when translating $4^{x+y} = 1$. Students sometimes hastily write $x+y = 1$ instead of $x+y = 0$, completely derailing the subsequent algebra. Always remember $a^0 = 1$. ### Final Answer Therefore, the correct answer is **$\frac{1}{2}$ and $-\frac{1}{2}$**.