1. **Problem statement:** Given that $\log_a 2 = 5$, find: (a) $\log_a 32$ (b) $\log_{\sqrt{a}} 2$ (c) The value of $a$ correct to 3 significant figures. 2. **Recall logarithm properties:** - $\log_a (xy) = \log_a x + \log_a y$ - $\log_a (x^k) = k \log_a x$ - Change of base formula: $\log_b x = \frac{\log_a x}{\log_a b}$ 3. **Part (a): Find $\log_a 32$** - Note that $32 = 2^5$ - Using the power rule: $$\log_a 32 = \log_a (2^5) = 5 \log_a 2$$ - Substitute $\log_a 2 = 5$: $$\log_a 32 = 5 \times 5 = 25$$ 4. **Part (b): Find $\log_{\sqrt{a}} 2$** - Recall $\sqrt{a} = a^{1/2}$ - Using change of base formula: $$\log_{\sqrt{a}} 2 = \frac{\log_a 2}{\log_a \sqrt{a}}$$ - Calculate denominator: $$\log_a \sqrt{a} = \log_a a^{1/2} = \frac{1}{2} \log_a a = \frac{1}{2} \times 1 = \frac{1}{2}$$ - Substitute values: $$\log_{\sqrt{a}} 2 = \frac{5}{\frac{1}{2}} = 5 \times 2 = 10$$ 5. **Part (c): Find $a$ given $\log_a 2 = 5$** - By definition: $$\log_a 2 = 5 \implies a^5 = 2$$ - Solve for $a$: $$a = \sqrt[5]{2} = 2^{\frac{1}{5}}$$ - Calculate numerical value: $$a \approx 2^{0.2} \approx 1.149$$ **Final answers:** - (a) $\log_a 32 = 25$ - (b) $\log_{\sqrt{a}} 2 = 10$ - (c) $a \approx 1.15$ (3 significant figures)