Here is the solution to the logarithmic expression. Step 1: Simplify the numerator using logarithm properties. The numerator is $\log \sqrt{27} + \log \sqrt{8} - \log \sqrt{125}$. First, rewrite the square roots as powers: $$\sqrt{27} = 27^{1/2}$$ $$\sqrt{8} = 8^{1/2}$$ $$\sqrt{125} = 125^{1/2}$$ Substitute these into the numerator: $$\log (27^{1/2}) + \log (8^{1/2}) - \log (125^{1/2})$$ Apply the logarithm power rule, $n \log a = \log a^n$: $$\frac{1}{2} \log 27 + \frac{1}{2} \log 8 - \frac{1}{2} \log 125$$ Factor out $\frac{1}{2}$: $$\frac{1}{2} (\log 27 + \log 8 - \log 125)$$ Rewrite the numbers inside the logarithms as powers of their prime factors: $$27 = 3^3$$ $$8 = 2^3$$ $$125 = 5^3$$ Substitute these back: $$\frac{1}{2} (\log (3^3) + \log (2^3) - \log (5^3))$$ Apply the logarithm power rule again: $$\frac{1}{2} (3 \log 3 + 3 \log 2 - 3 \log 5)$$ Factor out $3$: $$\frac{1}{2} \cdot 3 (\log 3 + \log 2 - \log 5)$$ $$\frac{3}{2} (\log 3 + \log 2 - \log 5)$$ Apply the logarithm product rule, $\log a + \log b = \log (ab)$, and the quotient rule, $\log a - \log b = \log \left(\frac{a}{b}\right)$: $$\frac{3}{2} (\log (3 \times 2) - \log 5)$$ $$\frac{3}{2} (\log 6 - \log 5)$$ Step 2: Simplify the denominator. The denominator is $\log 6 - \log 5$. This is already in a simplified form for comparison. Step 3: Combine the simplified numerator and denominator. The original expression becomes: $$\frac{\frac{3}{2} (\log 6 - \log 5)}{\log 6 - \log 5}$$ Since $(\log 6 - \log 5)$ is a common factor in both the numerator and the denominator, and it is not zero (because $\log 6 \neq \log 5$), we can cancel it out. $$ \frac{3}{2} $$ The final answer is $\boxed{\frac{3}{2}}$. Got more? Send 'em!