What is 4log_{6}(2)+log_{6}(3)+6log_{36}(3) ?

The solution to 4log_{6}(2)+log_{6}(3)+6log_{36}(3) is 1+3log_{6}(2)+6log_{36}(3)

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simplify 4log_{6}(2)+log_{6}(3)+6log_{36}(3)

simplify

4 lo g_{6} (2) + lo g_{6} (3) + 6 lo g_{36} (3)

1 + 3 lo g_{6} (2) + 6 lo g_{36} (3)

Decimal Notation

4

Solution steps

4 lo g_{6} (2) + lo g_{6} (3) + 6 lo g_{36} (3)

Apply log rule:

b \cdot lo g_{a} (x) = lo g_{a} (x^{b})

4 lo g_{6} (2) = lo g_{6} (2^{4})

= lo g_{6} (2^{4}) + lo g_{6} (3) + 6 lo g_{36} (3)

lo g_{6} (2^{4}) + lo g_{6} (3) = lo g_{6} (48)

= lo g_{6} (48) + 6 lo g_{36} (3)

lo g_{6} (48) = 1 + 3 lo g_{6} (2)

= 1 + 3 lo g_{6} (2) + 6 lo g_{36} (3)

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What is 4log_{6}(2)+log_{6}(3)+6log_{36}(3) ?
The solution to 4log_{6}(2)+log_{6}(3)+6log_{36}(3) is 1+3log_{6}(2)+6log_{36}(3)