What is log_{2}(12)+3log_{2}(3)-log_{2}(9) ?

The solution to log_{2}(12)+3log_{2}(3)-log_{2}(9) is 2log_{2}(6)

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simplify log_{2}(12)+3log_{2}(3)-log_{2}(9)

simplify

lo g_{2} (12) + 3 lo g_{2} (3) - lo g_{2} (9)

2 lo g_{2} (6)

Decimal Notation

5.16992 \dots

Solution steps

lo g_{2} (12) + 3 lo g_{2} (3) - lo g_{2} (9)

Apply log rule:

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

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

= lo g_{2} (12) + lo g_{2} (3^{3}) - lo g_{2} (9)

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

lo g_{2} (12) + lo g_{2} (3^{3}) = lo g_{2} (12 \cdot 3^{3})

= lo g_{2} (12 \cdot 3^{3}) - lo g_{2} (9)

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{2} (12 \cdot 3^{3}) - lo g_{2} (9) = lo g_{2} (\frac{12 \cdot 3 ^{3}}{9})

= lo g_{2} (\frac{12 \cdot 3 ^{3}}{9})

Cancel

\frac{12 \cdot 3 ^{3}}{9} : 36

= lo g_{2} (36)

Factor the number:

36 = 6^{2}

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

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), x > 0

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

= 2 lo g_{2} (6)

75 b^{3}

(x + 1)^{3} + (x - 1)^{3}

\frac{2 ^{2} \cdot 3 ^{3} \cdot 5 ^{7}}{3 ^{2} \cdot 5 ^{6} \cdot 2 ^{2}}

(2 x - 5) (4 x^{2} + 10 x + 25)

\frac{x}{cos ( x )}

What is log_{2}(12)+3log_{2}(3)-log_{2}(9) ?
The solution to log_{2}(12)+3log_{2}(3)-log_{2}(9) is 2log_{2}(6)