What is log_{3}(12)-log_{3}(28)+log_{3}(63) ?

The solution to log_{3}(12)-log_{3}(28)+log_{3}(63) is 3

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simplify log_{3}(12)-log_{3}(28)+log_{3}(63)

simplify

lo g_{3} (12) - lo g_{3} (28) + lo g_{3} (63)

3

Solution steps

lo g_{3} (12) - lo g_{3} (28) + lo g_{3} (63)

Apply log rule:

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

lo g_{3} (12) - lo g_{3} (28) = lo g_{3} (\frac{12}{28})

= lo g_{3} (\frac{12}{28}) + lo g_{3} (63)

Apply log rule:

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

lo g_{3} (\frac{12}{28}) + lo g_{3} (63) = lo g_{3} (\frac{12}{28} \cdot 63)

= lo g_{3} (\frac{12}{28} \cdot 63)

\frac{12}{28} \cdot 63 = 27

= lo g_{3} (27)

Factor the number:

27 = 3^{3}

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

Apply log rule:

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

= 3

(5 \cdot 3)^{3}

x^{\frac{1}{3}} \cdot x

x^{2} = - 121

43

6 (2 x - 1) < - 3 (5 - 4 x)

What is log_{3}(12)-log_{3}(28)+log_{3}(63) ?
The solution to log_{3}(12)-log_{3}(28)+log_{3}(63) is 3