[tex]\frac{log_2{20} }{log_2{12} }+ log_12{0,05}[/tex]

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To solve this expression, we first simplify the logarithms:

Since the base of both logarithms is 2, we can use the change of base formula to simplify this expression:

[tex]\frac{log_2{20}}{log2{12}} = log{12}{20}[/tex]

To simplify this expression, we can rewrite 0.05 as a fraction with a power of 10:

0.05 = 5 / 100 = 5 / (10^2) = 5 / (2^2 5^2) = 1 / (2^2 5)

Therefore, [tex]log{12}{0,05} = log{12}{(1 / (2^2 * 5))}[/tex]

Using the properties of logarithms, we can then rewrite this as:

[tex] = log{12}{1} - log{12}{(2^2 5)} [/tex]
[tex] = 0 - (log{12}{2^2} + log{12}{5}) [/tex]
[tex] = -(2 log{12}{2} + log{12}{5}) [/tex]

Putting it all back together:

[tex]log{12}{20} - (2 * log{12}{2} + log_{12}{5})[/tex]