Problems from Properties of the product and the quotient of the logarithms

Development:

First it is necessary to see if it is possible to decompose every number into powers of the same base as the logarithm base and then to apply the properties learned to solve them:

In the first case all the factors can be expressed as powers of $$5$$, so: $$$log_5 \dfrac{25\cdot 125\cdot \dfrac{1}{625}}{\dfrac{1}{5}}=log_5 \dfrac{5^2\cdot 5^3\cdot \dfrac{1}{5^4}}{\dfrac{1}{5}}=log_5 \dfrac{5^2\cdot 5^3\cdot 5^{-4}}{5^{-1}}$$$ Now that everything is expressed in powers of base $$5$$ we can apply the properties of the product, the quotient and the powers of the logarithms:

$$(log_5 5^2+log_5 5^3+log_5 5^{-4})-log_5 5^{-1}=(2+3-4)-(-1)=1+1=2$$

In the second case,we can express the numbers as powers of $$3$$:

$$$log_3 \dfrac{27\cdot\dfrac{1}{81}}{9\cdot\dfrac{1}{\sqrt{3}}}= log_3 \dfrac{3^3\cdot\dfrac{1}{3^4}}{3^2\cdot\dfrac{1}{3^{\frac{1}{2}}}}= log_3 \dfrac{3^3\cdot3^{-4}}{3^2\cdot3^{-\frac{1}{2}}}=$$$ $$$=(log_3 3^3+log_3 3^{-4})-(log_3 3^2+log_3 3^{-\frac{1}{2}})=$$$ $$$=(3-4)-(2-\dfrac{1}{2})=-\dfrac{7}{2}$$$

In the third case, it is necessary to use the powers of $$10$$: $$$log\dfrac{10.000\cdot0,1}{0,001\cdot100}=log\dfrac{10^4\cdot10^{-1}}{10^{-3}\cdot10^2}=$$$ $$$=(log10^4+log10^{-1})-(log10^{-3}+log10^2)=$$$ $$$=(4-1)-(-3+2)=3+1=4$$$

Finally, in the last case, we can try to obtain as much powers of base $$2$$ as possible:

$$$log_2 \dfrac{12}{\sqrt{36}\cdot 16}=log_2 \dfrac{3\cdot4}{6\cdot 2^4}=log_2 \dfrac{3\cdot2^2}{2\cdot3\cdot 2^4}=$$$ $$$=(log_2 3+log_2 2^2)-(log_2 2+log_2 3+log_2 2^4)=$$$ $$$=log_2 3+2-1-log_2 3-4=-3$$$

Solution:

$$2, -\dfrac{7}{2}, 4, -3$$

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