Question: Q4.11
1 Point
log_2(\sqrt{2}) =
1
1/2
-1/2
2
-1
Q4.12
1 Point
If 3^{2x} = 9, then x =
3
1
3/2
6
Q4.13
1 Point
top-right: browser tabs and header text; center: math questions Q4.11 and Q4.12 with multiple-choice circles; bottom-left: sidebar navigation; bottom-right: empty page area and laptop keyboard visible.User: answer both questions
1. **Problem 1: Calculate $\log_2(\sqrt{2})$.**
2. Recall the property of logarithms: $\log_b(a^c) = c \log_b(a)$.
3. Since $\sqrt{2} = 2^{1/2}$, we have:
$$\log_2(\sqrt{2}) = \log_2(2^{1/2}) = \frac{1}{2} \log_2(2)$$
4. We know $\log_2(2) = 1$, so:
$$\log_2(\sqrt{2}) = \frac{1}{2} \times 1 = \frac{1}{2}$$
5. **Answer for Q4.11 is $\frac{1}{2}$.**
---
6. **Problem 2: Solve for $x$ in $3^{2x} = 9$.**
7. Express $9$ as a power of $3$: $9 = 3^2$.
8. So the equation becomes:
$$3^{2x} = 3^2$$
9. Since the bases are equal, set the exponents equal:
$$2x = 2$$
10. Divide both sides by 2:
$$\cancel{2}x = \cancel{2}$$
$$x = 1$$
11. **Answer for Q4.12 is $1$.**