What is log_{2}(6)-log_{2}(15)+log_{2}(20) ?

The solution to log_{2}(6)-log_{2}(15)+log_{2}(20) is 3

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simplify log_{2}(6)-log_{2}(15)+log_{2}(20)

simplify

lo g_{2} (6) - lo g_{2} (15) + lo g_{2} (20)

3

Solution steps

lo g_{2} (6) - lo g_{2} (15) + lo g_{2} (20)

Apply log rule:

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

lo g_{2} (6) - lo g_{2} (15) = lo g_{2} (\frac{6}{15})

= lo g_{2} (\frac{6}{15}) + lo g_{2} (20)

Apply log rule:

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

lo g_{2} (\frac{6}{15}) + lo g_{2} (20) = lo g_{2} (\frac{6}{15} \cdot 20)

= lo g_{2} (\frac{6}{15} \cdot 20)

\frac{6}{15} \cdot 20 = 8

= lo g_{2} (8)

Factor the number:

8 = 2^{3}

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

Apply log rule:

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

= 3

9 t^{2} + 24 t

2 (x - 1) (x + 4) + (x - 1)^{2} > 0

x^{2} - 4 x + 7 x = 0

x^{3} < 5

(x + 6) (x - 6) - (x - 6) (x - 6)

What is log_{2}(6)-log_{2}(15)+log_{2}(20) ?
The solution to log_{2}(6)-log_{2}(15)+log_{2}(20) is 3