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

The solution to log_{6}(2)-log_{6}(4)+log_{6}(3) is log_{6}(3/2)

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

simplify

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

lo g_{6} (\frac{3}{2})

Decimal Notation

0.22629 \dots

Solution steps

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

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{6} (3) - lo g_{6} (4) = lo g_{6} (\frac{3}{4})

= lo g_{6} (2) + lo g_{6} (\frac{3}{4})

Apply log rule:

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

lo g_{6} (2) + lo g_{6} (\frac{3}{4}) = lo g_{6} (2 \cdot \frac{3}{4})

= lo g_{6} (2 \cdot \frac{3}{4})

2 \cdot \frac{3}{4} = \frac{3}{2}

= lo g_{6} (\frac{3}{2})

6 \cdot ln (\frac{6}{9.1161})

\frac{1}{3} ln (\frac{3 x + 16 + 9 x ^{2}}{4})

ln (\frac{12}{12 + d})

2 - 2 ln (e + 1) + ln (4)

lo g_{2} (\frac{8 d ^{4}}{n ^{3}})

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