What is log_{2}(2/3)+log_{2}(4/5)-log_{2}(1/30) ?

The solution to log_{2}(2/3)+log_{2}(4/5)-log_{2}(1/30) is 4

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simplify log_{2}(2/3)+log_{2}(4/5)-log_{2}(1/30)

simplify

lo g_{2} (\frac{2}{3}) + lo g_{2} (\frac{4}{5}) - lo g_{2} (\frac{1}{30})

4

Solution steps

lo g_{2} (\frac{2}{3}) + lo g_{2} (\frac{4}{5}) - lo g_{2} (\frac{1}{30})

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

lo g_{2} (\frac{2}{3}) + lo g_{2} (\frac{4}{5}) = lo g_{2} (\frac{2}{3} \cdot \frac{4}{5})

= lo g_{2} (\frac{2}{3} \cdot \frac{4}{5}) - lo g_{2} (\frac{1}{30})

\frac{2}{3} \cdot \frac{4}{5} = \frac{8}{15}

= lo g_{2} (\frac{8}{15}) - lo g_{2} (\frac{1}{30})

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{2} (\frac{8}{15}) - lo g_{2} (\frac{1}{30}) = lo g_{2} (\frac{\frac{8}{15}}{\frac{1}{30}})

= lo g_{2} (\frac{\frac{8}{15}}{\frac{1}{30}})

\frac{\frac{8}{15}}{\frac{1}{30}} = 16

= lo g_{2} (16)

Rewrite

16

in power-base form:

16 = 2^{4}

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

Apply log rule:

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

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

= 4 lo g_{2} (2)

Apply log rule:

lo g_{a} (a) = 1

lo g_{2} (2) = 1

= 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 4

(21 - 27)^{2}

lo g_{4} (24) + lo g_{4} (12) - lo g_{4} (18)

(6 y^{2} + 5 y^{4})^{2}

- x + 2 x

ln (2^{x})

What is log_{2}(2/3)+log_{2}(4/5)-log_{2}(1/30) ?
The solution to log_{2}(2/3)+log_{2}(4/5)-log_{2}(1/30) is 4