What is log_{3}(6c)+log_{3}(1/12) ?

The solution to log_{3}(6c)+log_{3}(1/12) is log_{3}(c/2)

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simplify log_{3}(6c)+log_{3}(1/12)

simplify

lo g_{3} (6 c) + lo g_{3} (\frac{1}{12})

lo g_{3} (\frac{c}{2})

Solution steps

lo g_{3} (6 c) + lo g_{3} (\frac{1}{12})

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{3} (6 c) + lo g_{3} (\frac{1}{12}) = lo g_{3} (6 \cdot \frac{1}{12} c)

= lo g_{3} (6 \cdot \frac{1}{12} c)

Simplify

6 c \frac{1}{12} : \frac{c}{2}

= lo g_{3} (\frac{c}{2})

- \frac{1}{45} ln (\frac{18 - ( - 7 )}{22 - ( - 7 )})

6000 ln (100) - 6000 ln (50)

lo g_{3} (147) - lo g_{3} (70) - lo g_{3} (189)

ln (2 \cdot 1)

x - ln ∣ x + 4 ∣ + ln ∣ x ∣

What is log_{3}(6c)+log_{3}(1/12) ?
The solution to log_{3}(6c)+log_{3}(1/12) is log_{3}(c/2)