1. **State the problem:** We need to solve the equation $$3^x - 3^y = 234$$ for the variables $x$ and $y$. 2. **Analyze the equation:** The equation involves exponential terms with the same base 3 but different exponents $x$ and $y$. 3. **Rewrite the equation:** Let’s express $3^x$ in terms of $3^y$: $$3^x = 3^y + 234$$ 4. **Introduce a substitution:** Let $a = 3^y$. Then the equation becomes: $$3^x = a + 234$$ Since $3^x = 3^{y + (x - y)} = 3^y \cdot 3^{x - y} = a \cdot 3^{x - y}$$, we can write: $$a \cdot 3^{x - y} = a + 234$$ 5. **Divide both sides by $a$ (assuming $a \neq 0$):** $$3^{x - y} = 1 + \frac{234}{a}$$ 6. **Set $k = x - y$ and rewrite:** $$3^k = 1 + \frac{234}{a}$$ 7. **Since $a = 3^y$, $a$ is a power of 3. We look for integer powers of 3 that divide 234 nicely.** Try $a = 3^1 = 3$: $$3^k = 1 + \frac{234}{3} = 1 + 78 = 79$$ 79 is not a power of 3. Try $a = 3^2 = 9$: $$3^k = 1 + \frac{234}{9} = 1 + 26 = 27$$ 27 is $3^3$, so $k = 3$. 8. **From $a = 3^y = 9$, we get $y = 2$. From $k = x - y = 3$, we get $x = y + 3 = 5$.** 9. **Check the solution:** $$3^5 - 3^2 = 243 - 9 = 234$$ Correct. **Final answer:** $$x = 5, \quad y = 2$$