Logarithm Expressions Ec89D1

1. **Problem statement:** Given $\log_a y = m$, express each of the following in terms of $m$: a) $\log_y a$ b) $\log_{\sqrt{y}} a$ c) $\log_{1/a} y$ d) $\log_{1/y} \sqrt{a}$ 2. **Recall the change of base formula:** $$\log_b c = \frac{\log_a c}{\log_a b}$$ This formula allows us to rewrite logarithms with any base in terms of logarithms with base $a$. 3. **Given:** $$\log_a y = m$$ From this, we can also write: $$\log_a a = 1$$ 4. **Part (a):** Find $\log_y a$ Using change of base: $$\log_y a = \frac{\log_a a}{\log_a y} = \frac{1}{m}$$ 5. **Part (b):** Find $\log_{\sqrt{y}} a$ Note that $\sqrt{y} = y^{1/2}$, so: $$\log_a \sqrt{y} = \log_a y^{1/2} = \frac{1}{2} \log_a y = \frac{m}{2}$$ Then, $$\log_{\sqrt{y}} a = \frac{\log_a a}{\log_a \sqrt{y}} = \frac{1}{m/2} = \frac{2}{m}$$ 6. **Part (c):** Find $\log_{1/a} y$ Note that $1/a = a^{-1}$, so: $$\log_a (1/a) = \log_a a^{-1} = -1$$ Then, $$\log_{1/a} y = \frac{\log_a y}{\log_a (1/a)} = \frac{m}{-1} = -m$$ 7. **Part (d):** Find $\log_{1/y} \sqrt{a}$ Note that $1/y = y^{-1}$ and $\sqrt{a} = a^{1/2}$, so: $$\log_a (1/y) = \log_a y^{-1} = -m$$ $$\log_a \sqrt{a} = \log_a a^{1/2} = \frac{1}{2}$$ Then, $$\log_{1/y} \sqrt{a} = \frac{\log_a \sqrt{a}}{\log_a (1/y)} = \frac{1/2}{-m} = -\frac{1}{2m}$$ **Final answers:** $$\text{a) } \log_y a = \frac{1}{m}$$ $$\text{b) } \log_{\sqrt{y}} a = \frac{2}{m}$$ $$\text{c) } \log_{1/a} y = -m$$ $$\text{d) } \log_{1/y} \sqrt{a} = -\frac{1}{2m}$$