# Evaluate $\dfrac{\log \sqrt{27} + \log \sqrt{8} -\log \sqrt{125}}{\log 6 -\log 5}$

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.

###### Step: 1

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

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

###### Step: 2

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.

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

###### Step: 3

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

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

###### Step: 4

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.

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

###### Step: 5

It can be expressed as two multiplying factors.

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

###### Step: 6

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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