How calculate log base5 (12)+log base5 (0,12)-2log base5 (6)?

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You need to use the power property of logarithms for the term `2log_5 6` , such that:

`2log_5 6 = log_5 6^2 => 2log_5 6 = log_5 36`

You need to convert the summation of the first two logarithms into the logarithm of product, such that:

`log_5 12 + log_5 (0.12) = log_5 (12*(0.12))`

`log_5 12 + log_5 (0.12) = log_5 (12*(12/10))`

Reducing duplicate factors yields:

`log_5 12 + log_5 (0.12) = log_5 (144/10)`

Replacing `log_5 (144/10)` for `log_5 12 + log_5 (0.12)` yields:

`log_5 12 + log_5 (0.12) - log_5 36 =log_5 (144/10) - log_5 36 `

You need to convert the difference of logarithms into the logarithm of quotient,such that:

`log_5 12 + log_5 (0.12) - log_5 36 = log_5 (144/(10*36)))`

`log_5 12 + log_5 (0.12) - log_5 36 = log_5 ((12*12)/(10*36))`

`log_5 12 + log_5 (0.12) - log_5 36 = log_5 (6/5*6/18)`

`log_5 12 + log_5 (0.12) - log_5 36 = log_5 (6/15)`

Hence, evaluating the given expression, using the properties of logarithms, yields `log_5 12 + log_5 (0.12) - log_5 36 = log_5 (6/15).`

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