Five logarithmic terms formed a mathematical relation in division form. logarithms of square roots of $27$, $8$ and $125$, and also logarithms of $6$ and $5$ are the numbers involved in forming this mathematical expression.

The logarithm of square root of $125$ is subtracted from sum of the logarithm of square root of $27$ and log of square root of $8$, to form the numerator. Similarly, the log of $5$ is subtracted from log of $6$ to form the denominator. The value of ratio of them is required to evaluate in this problem.

The radicand of each logarithmic term in the numerator can be converted as cube of a number.

$=$ $\dfrac{\log \sqrt{3^3} + \log \sqrt{2^3} -\log \sqrt{5^3}}{\log 6 -\log 5}$

The symbol square root ($\sqrt{\,\,\,\,}$) represents an exponent $\dfrac{1}{2}$.

$=$ $\dfrac{\log {\Big(3^3\Big)}^{\frac{1}{2}} + \log {\Big(2^3\Big)}^{\frac{1}{2}} -\log {\Big(5^3\Big)}^{\frac{1}{2}}}{\log 6 -\log 5}$

Use power rule of exponents to simplify power of an exponential term.

$=$ $\dfrac{\log {(3)}^{\frac{3}{2}} + \log {(2)}^{\frac{3}{2}} -\log {(5)}^{\frac{3}{2}}}{\log 6 -\log 5}$

The number of the logarithm term is in exponential notation. So, apply power rule of the logarithms to simplify each term in the numerator.

$=$ $\dfrac{\dfrac{3}{2} \, (\log 3) + \dfrac{3}{2} \, (\log 2) -\dfrac{3}{2} \, (\log 5)}{\log 6 -\log 5}$

The rational number $\dfrac{3}{2}$ is a common multiplying factor of each logarithmic term in numerator. So, take $\dfrac{3}{2}$ common from three terms.

$=$ $\dfrac{\dfrac{3}{2} \Big[\log 3 + \log 2 -\log 5 \Big]}{\log 6 -\log 5}$

It can be expressed as two multiplying factors.

$=$ $\dfrac{3}{2} \times \dfrac{\log 3 + \log 2 -\log 5}{\log 6 -\log 5}$

Observe the expression in numerator and denominator carefully. The numerator can be expressed same as the expression in the denominator by applying product rule of logarithms to addition of $\log 3$ and $\log 2$ terms.

$=$ $\dfrac{3}{2} \times \dfrac{\log (3 \times 2) -\log 5}{\log 6 -\log 5}$

$=$ $\dfrac{3}{2} \times \dfrac{\log 6 -\log 5}{\log 6 -\log 5}$

$=$ $\require{cancel} \dfrac{3}{2} \times \dfrac{\cancel{\log 6 -\log 5}}{\cancel{\log 6 -\log 5}}$

$=$ $\dfrac{3}{2} \times 1$

$\therefore \,\,\,\,\,\, \dfrac{\log \sqrt{27} + \log \sqrt{8} -\log \sqrt{125}}{\log 6 -\log 5}$ $=$ $\dfrac{3}{2}$

Therefore, the answer for this problem is $\dfrac{3}{2}$ and it is required solution for this logarithm problem mathematically.

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