What is 1/2 [1+log_{2}(a)-log_{2}(b)-log_{2}(3)] ?

The solution to 1/2 [1+log_{2}(a)-log_{2}(b)-log_{2}(3)] is (1+log_{2}(a/(3b)))/2

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simplify 1/2 [1+log_{2}(a)-log_{2}(b)-log_{2}(3)]

simplify

\frac{1}{2} [1 + lo g_{2} (a) - lo g_{2} (b) - lo g_{2} (3)]

\frac{1 + lo g _{2} ( \frac{a}{3 b} )}{2}

Solution steps

\frac{1}{2} (1 + lo g_{2} (a) - lo g_{2} (b) - lo g_{2} (3))

Simplify

lo g_{2} (a) - lo g_{2} (b) - lo g_{2} (3) : lo g_{2} (\frac{a}{b \cdot 3})

= \frac{1}{2} (1 + lo g_{2} (\frac{a}{b \cdot 3}))

Expand

\frac{1}{2} (1 + lo g_{2} (\frac{a}{b \cdot 3})) : \frac{1 + lo g _{2} ( \frac{a}{3 b} )}{2}

= \frac{1 + lo g _{2} ( \frac{a}{3 b} )}{2}

lo g_{a} (\frac{x ^{2} y}{z})

ln (\frac{x}{3 y})

lo g_{10} (9) + lo g_{10} (5) - lo g_{10} (15)

lo g_{10} (\frac{( a ^{6} b ^{3} )}{c ^{5}})

What is 1/2 [1+log_{2}(a)-log_{2}(b)-log_{2}(3)] ?
The solution to 1/2 [1+log_{2}(a)-log_{2}(b)-log_{2}(3)] is (1+log_{2}(a/(3b)))/2