What is log_{8}(60)-log_{8}(3)-log_{8}(5) ?

The solution to log_{8}(60)-log_{8}(3)-log_{8}(5) is 2/3

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simplify log_{8}(60)-log_{8}(3)-log_{8}(5)

simplify

lo g_{8} (60) - lo g_{8} (3) - lo g_{8} (5)

\frac{2}{3}

Solution steps

lo g_{8} (60) - lo g_{8} (3) - lo g_{8} (5)

Apply log rule:

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

lo g_{8} (60) - lo g_{8} (3) = lo g_{8} (\frac{60}{3})

= lo g_{8} (\frac{60}{3}) - lo g_{8} (5)

Apply log rule:

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

lo g_{8} (\frac{60}{3}) - lo g_{8} (5) = lo g_{8} (\frac{\frac{60}{3}}{5})

= lo g_{8} (\frac{\frac{60}{3}}{5})

Factor the number:

8 = 2^{3}

= lo g_{2^{3}} (\frac{\frac{60}{3}}{5})

Apply log rule:

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

lo g_{2^{3}} (\frac{\frac{60}{3}}{5}) = \frac{1}{3} lo g_{2} (\frac{\frac{60}{3}}{5})

= \frac{1}{3} lo g_{2} (\frac{\frac{60}{3}}{5})

\frac{\frac{60}{3}}{5} = 2^{2}

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

Apply log rule:

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

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

= \frac{1}{3} \cdot 2

\frac{1}{3} \cdot 2 = \frac{2}{3}

= \frac{2}{3}

10 q - 15 = 5 (2 q + 4)

3364 g^{2} - 10556 g + 8281

x^{2} + 3 x - 1 < 3

0.0657 - (0.7 5^{2} \times 0.0676)

2 x^{2} = 100

What is log_{8}(60)-log_{8}(3)-log_{8}(5) ?
The solution to log_{8}(60)-log_{8}(3)-log_{8}(5) is 2/3