What is log_{3}(1/2)-log_{9}(1/2) ?

The solution to log_{3}(1/2)-log_{9}(1/2) is -log_{3}(2)+log_{9}(2)

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simplify log_{3}(1/2)-log_{9}(1/2)

simplify

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

- lo g_{3} (2) + lo g_{9} (2)

Decimal Notation

- 0.31546 \dots

Solution steps

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

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

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

= - lo g_{3} (2) - (- lo g_{9} (2))

Apply rule:

a - (- b) = a + b

- lo g_{3} (2) - (- lo g_{9} (2)) = - lo g_{3} (2) + lo g_{9} (2)

= - lo g_{3} (2) + lo g_{9} (2)

lo g_{2} (\frac{i}{j s})

ln (\frac{u}{3} + 1) - ln (\frac{u}{3} - 1)

lo g_{7} (9 x)

lo g_{5} (\frac{x y}{z})

lo g_{10} (y) - 15 lo g_{10} (z)

What is log_{3}(1/2)-log_{9}(1/2) ?
The solution to log_{3}(1/2)-log_{9}(1/2) is -log_{3}(2)+log_{9}(2)