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

The solution to 3log_{3}(2)+2log_{3}(4)-log_{3}(6) is log_{3}(64/3)

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

simplify

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

lo g_{3} (\frac{64}{3})

Decimal Notation

2.78557 \dots

Solution steps

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 lo g_{3} (4) = lo g_{3} (4^{2})

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

Apply log rule:

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

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

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

\frac{4 ^{2}}{6} = \frac{8}{3}

= 3 lo g_{3} (2) + lo g_{3} (\frac{8}{3})

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

3 lo g_{3} (2) = lo g_{3} (2^{3})

= lo g_{3} (2^{3}) + lo g_{3} (\frac{8}{3})

Apply log rule:

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

lo g_{3} (2^{3}) + lo g_{3} (\frac{8}{3}) = lo g_{3} (2^{3} \cdot \frac{8}{3})

= lo g_{3} (2^{3} \cdot \frac{8}{3})

2^{3} \cdot \frac{8}{3} = \frac{64}{3}

= lo g_{3} (\frac{64}{3})

2 lo g_{b} (x) + 9 lo g_{b} (z)

lo g_{5} (\frac{2 x - 1}{5})

4 ln (x) + 3 ln (y) - 6 ln (z)

lo g_{10} (\frac{x ^{3}}{z ^{2} y})

ln (w) + 3 ln (4 z)

What is 3log_{3}(2)+2log_{3}(4)-log_{3}(6) ?
The solution to 3log_{3}(2)+2log_{3}(4)-log_{3}(6) is log_{3}(64/3)