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

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

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

simplify

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

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

Solution steps

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

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

\frac{1}{9} (lo g_{2} (z) + 4 lo g_{2} (y))

\frac{3 lo g _{10} ( 2 )}{lo g _{10} ( 2 ) - lo g _{10} ( 3 )}

lo g_{10} (z^{6} y)

lo g_{2} (3 x y^{2})

lo g_{10} (8 k)

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