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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

4 [lo g_{2} (k) - lo g_{2} (k - t)]

ln (28) 0.125 + ln (\frac{3}{5}) 0.6 - 49 ln (7 e^{3})

lo g_{4} (\frac{x - 2}{x ^{5}})

lo g_{10} (\frac{1 0 ^{2} \cdot 1000}{x ^{3} \cdot y ^{- 2}})

lo g_{10} (\frac{10}{a ^{2} + b ^{2}})

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