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

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

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

simplify

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

2 lo g_{2} (3)

Decimal Notation

3.16992 \dots

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Cancel

\frac{12 \cdot 3}{2 ^{2}} : 9

= lo g_{2} (9)

Factor the number:

9 = 3^{2}

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

Apply log rule:

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

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

= 2 lo g_{2} (3)

lo g_{10} (10 x^{5})

lo g_{5} (25 z)

- \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + C

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

ln (\frac{9}{19})

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