What is 3log_{4}(3)-log_{4}(12)+2 ?

The solution to 3log_{4}(3)-log_{4}(12)+2 is log_{2}(3)+1

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simplify 3log_{4}(3)-log_{4}(12)+2

simplify

3 lo g_{4} (3) - lo g_{4} (12) + 2

lo g_{2} (3) + 1

Decimal Notation

2.58496 \dots

Solution steps

3 lo g_{4} (3) - lo g_{4} (12) + 2

Apply log rule:

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

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

= lo g_{4} (3^{3}) - lo g_{4} (12) + 2

lo g_{4} (3^{3}) - lo g_{4} (12) = lo g_{4} (\frac{9}{4})

= lo g_{4} (\frac{9}{4}) + 2

lo g_{4} (\frac{9}{4}) = lo g_{2} (3) - 1

= (lo g_{2} (3) - 1) + 2

Apply rule:

(a) = a

(lo g_{2} (3) - 1) = lo g_{2} (3) - 1

= lo g_{2} (3) - 1 + 2

Add the numbers:

- 1 + 2 = 1

= lo g_{2} (3) + 1

2 (\frac{1}{2} x^{2} + 8)^{2} - 8

\frac{21}{60}

x^{4} - 10 x^{2} y^{2} + 25 y^{4}

\frac{1}{3 + 2}

\frac{7 x}{x ^{2} - 4} + \frac{2}{x + 2}

What is 3log_{4}(3)-log_{4}(12)+2 ?
The solution to 3log_{4}(3)-log_{4}(12)+2 is log_{2}(3)+1