# How do you evaluate ((log_3 (1/27)) - (log_16 4))/((log_5 9)(log_5 125))?

$\frac{{\log}_{3} \left(\frac{1}{27}\right) - {\log}_{16} 4}{\left({\log}_{5} 9\right) \left({\log}_{5} 125\right)} = - {\log}_{10} \frac{78125}{\log} _ 10 531441$

$= - 0.85456788$

#### Explanation:

from the given

$\frac{{\log}_{3} \left(\frac{1}{27}\right) - {\log}_{16} 4}{\left({\log}_{5} 9\right) \left({\log}_{5} 125\right)}$

Let us do this in detail

${\log}_{3} \left(\frac{1}{27}\right) = {\log}_{3} \left(\frac{1}{3} ^ 3\right) = {\log}_{3} \left({3}^{-} 3\right) = - 3$

log_16 4=log_16 4^1=log_16 4^(2/2)=log_16 (4^2)^(1/2) =log_16 (16)^(1/2)=1/2

${\log}_{5} 125 = {\log}_{5} {5}^{3} = 3$

Now let us go back to the given

$\frac{{\log}_{3} \left(\frac{1}{27}\right) - {\log}_{16} 4}{\left({\log}_{5} 9\right) \left({\log}_{5} 125\right)} = \frac{- 3 - \frac{1}{2}}{\left({\log}_{5} 9\right) \cdot 3} = \frac{- \frac{7}{2}}{3 \cdot {\log}_{5} 9} = \frac{- \frac{7}{2}}{{\log}_{5} 729} = \frac{- \frac{7}{2}}{{\log}_{10} \frac{729}{\log} _ 10 5} = \left(- \frac{7}{2}\right) \cdot \left({\log}_{10} \frac{5}{\log} _ 10 729\right) = - {\log}_{10} \frac{78125}{\log} _ 10 531441$

$= - 0.85456788$

God bless .....I hope the explanation is useful.