Let's solve the first question step by step! 🎉 **Problem:** Solve for $x$ if $x + \log_2(2^x - 6) = 4$ 1. Imagine you have $x$ apples 🍎. 2. The expression $\log_2(2^x - 6)$ means "What power of 2 gives you $2^x - 6$?" 3. Let’s call $y = 2^x - 6$. 4. So, $\log_2(y)$ is the power to get $y$ from 2. 5. The equation is: $x + \log_2(y) = 4$ 6. Since $y = 2^x - 6$, rewrite $\log_2(y)$ as $\log_2(2^x - 6)$. 7. Let's try values for $x$ that make $2^x - 6$ positive (because log only works with positive numbers): **Try $x=3$:** - $2^3 - 6 = 8 - 6 = 2$ - $\log_2(2) = 1$ - $x + \log_2(2^x - 6) = 3 + 1 = 4$ ✔️ Great! This works! So, $x = 3$ is the answer! 🎯 --- Now for the second question: **Problem:** If $263 + 441 = 714$ in some base $b$, find $b$. 1. Imagine numbers in base $b$, where digits go from 0 to $b-1$. 2. Convert $263_b + 441_b = 714_b$ to decimals: - $263_b = 2b^2 + 6b + 3$ - $441_b = 4b^2 + 4b + 1$ - $714_b = 7b^2 + b + 4$ 3. Write the equation: $$2b^2 + 6b + 3 + 4b^2 + 4b + 1 = 7b^2 + b + 4$$ 4. Combine left side: $$6b^2 + 10b + 4 = 7b^2 + b + 4$$ 5. Move all terms to one side: $$6b^2 + 10b + 4 - 7b^2 - b - 4 = 0$$ 6. Simplify: $$-b^2 + 9b = 0$$ 7. Factor: $$b(-b + 9) = 0$$ 8. So, $b=0$ (not possible) or $b=9$. Answer: The base is 9! 🎉 Great job solving both problems! 🌟