Question:

Step 1: Express $a$ and $b$ in terms of basic logarithms. We are given that $a = \log 25$ and $b = \log 225$ (all logarithms are base 10). Let us decompose these: $$a = \log 25 = \log 5^2 = 2\log 5$$ $$b = \log 225 = \log(9 \times 25) = \log 9 + \log 25 = 2\log 3 + 2\log 5$$ Step 2: Find expressions for $\log 3$ and $\log 5$ in terms of $a$ and $b$. From the expressions above: $$b - a = 2\log 3 + 2\log 5 - 2\log 5 = 2\log 3$$ Therefore: $$\log 3 = \frac{b-a}{2}$$ Also, from Step 1: $$\log 5 = \frac{a}{2}$$ Step 3: Rewrite the given expression using logarithm properties. We need to find: $$\log\left(\frac{1}{81}\right) + \log\left(\frac{1}{2250}\right)$$ Using the property $\log\left(\frac{1}{x}\right) = -\log x$: $$\log\left(\frac{1}{81}\right) + \log\left(\frac{1}{2250}\right) = -\log 81 - \log 2250 = -(\log 81 + \log 2250)$$ Step 4: Calculate $\log 81$ in terms of $a$ and $b$. $$\log 81 = \log 3^4 = 4\log 3 = 4 \cdot \frac{b-a}{2} = 2(b-a) = 2b - 2a$$ Step 5: Calculate $\log 2250$ in terms of $a$ and $b$. First, factor 2250: $$2250 = 2 \times 9 \times 125 = 2 \times 3^2 \times 5^3$$ Therefore: $$\log 2250 = \log 2 + 2\log 3 + 3\log 5$$ We need to find $\log 2$. Using the fact that $10 = 2 \times 5$: $$\log 10 = 1 = \log 2 + \log 5$$ $$\log 2 = 1 - \log 5 = 1 - \frac{a}{2}$$ Step 6: Substitute to find $\log 2250$. $$\log 2250 = \log 2 + 2\log 3 + 3\log 5$$ $$= \left(1 - \frac{a}{2}\right) + 2 \cdot \frac{b-a}{2} + 3 \cdot \frac{a}{2}$$ $$= 1 - \frac{a}{2} + (b-a) + \frac{3a}{2}$$ $$= 1 + b - \frac{a}{2} - a + \frac{3a}{2}$$ $$= 1 + b + \left(-\frac{a}{2} - \frac{2a}{2} + \frac{3a}{2}\right)$$ $$= 1 + b + 0 = 1 + b$$ Step 7: Calculate $\log 81 + \log 2250$. $$\log 81 + \log 2250 = (2b - 2a) + (1 + b) = 3b - 2a + 1$$ Step 8: Find the final answer. $$\log\left(\frac{1}{81}\right) + \log\left(\frac{1}{2250}\right) = -(\log 81 + \log 2250)$$ $$= -(3b - 2a + 1)$$ $$= 2a - 3b - 1$$ **Final Answer:** The expression $\log\left(\dfrac{1}{81}\right) + \log\left(\dfrac{1}{2250}\right) = 2a - 3b - 1$ This matches **Option 3**.