What is log_{8}(20)-log_{8}(15)+log_{8}(12) ?

The solution to log_{8}(20)-log_{8}(15)+log_{8}(12) is 4/3

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simplify log_{8}(20)-log_{8}(15)+log_{8}(12)

simplify

lo g_{8} (20) - lo g_{8} (15) + lo g_{8} (12)

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

lo g_{8} (20) - lo g_{8} (15) + lo g_{8} (12)

Apply log rule:

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

lo g_{8} (20) - lo g_{8} (15) = lo g_{8} (\frac{20}{15})

= lo g_{8} (\frac{20}{15}) + lo g_{8} (12)

Cancel

\frac{20}{15} : \frac{4}{3}

= lo g_{8} (\frac{4}{3}) + lo g_{8} (12)

Apply log rule:

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

lo g_{8} (\frac{4}{3}) + lo g_{8} (12) = lo g_{8} (\frac{4}{3} \cdot 12)

= lo g_{8} (\frac{4}{3} \cdot 12)

Factor the number:

8 = 2^{3}

= lo g_{2^{3}} (\frac{4}{3} \cdot 12)

Apply log rule:

lo g_{a^{b}} (x) = \frac{1}{b} lo g_{a} (x), a > 0

lo g_{2^{3}} (\frac{4}{3} \cdot 12) = \frac{1}{3} lo g_{2} (\frac{4}{3} \cdot 12)

= \frac{1}{3} lo g_{2} (\frac{4}{3} \cdot 12)

\frac{4}{3} \cdot 12 = 16

= \frac{1}{3} lo g_{2} (16)

Factor the number:

16 = 2^{4}

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

Apply log rule:

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

lo g_{2} (2^{4}) = 4

= \frac{1}{3} \cdot 4

\frac{1}{3} \cdot 4 = \frac{4}{3}

= \frac{4}{3}

(x - 6) (x + 2)

x + x

(a - b)^{4}

3 \times \frac{1}{2}

tan (x) + \frac{cos ( x )}{1 + sin ( x )}

What is log_{8}(20)-log_{8}(15)+log_{8}(12) ?
The solution to log_{8}(20)-log_{8}(15)+log_{8}(12) is 4/3