1. The problem is to solve the equation $3^{x+3} = 3^x + 234$ for $x$. 2. Recall the properties of exponents: $3^{x+3} = 3^x \cdot 3^3 = 3^x \cdot 27$. 3. Substitute this into the equation: $$3^x \cdot 27 = 3^x + 234$$ 4. Rearrange to isolate terms: $$27 \cdot 3^x - 3^x = 234$$ 5. Factor out $3^x$: $$3^x (27 - 1) = 234$$ 6. Simplify inside the parentheses: $$3^x \cdot 26 = 234$$ 7. Divide both sides by 26: $$\cancel{26} \cdot 3^x = \frac{234}{\cancel{26}}$$ $$3^x = 9$$ 8. Recognize that $9 = 3^2$, so: $$3^x = 3^2$$ 9. Since the bases are equal and positive (and not 1), the exponents must be equal: $$x = 2$$ Final answer: $x = 2$