What is log_{2}(2x)-log_{2}(x+1) ?

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

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simplify log_{2}(2x)-log_{2}(x+1)

simplify

lo g_{2} (2 x) - lo g_{2} (x + 1)

1 + lo g_{2} (\frac{x}{x + 1})

Solution steps

lo g_{2} (2 x) - lo g_{2} (x + 1)

Simplify

lo g_{2} (2 x) : 1 + lo g_{2} (x)

= (1 + lo g_{2} (x)) - lo g_{2} (x + 1)

Apply rule:

(a) = a

(1 + lo g_{2} (x)) = 1 + lo g_{2} (x)

= 1 + lo g_{2} (x) - lo g_{2} (x + 1)

Apply log rule:

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

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

= 1 + lo g_{2} (\frac{x}{x + 1})

ln (\frac{e}{4})

lo g_{m} (m x^{2})

ln (2) + x ln (x) + x - 2 ln (x - 1) - 3

ln (\frac{x ^{2} x ^{2} + 7}{( x + 7 ) ^{9}})

lo g_{10} (\frac{x ^{4} y}{z})

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