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

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

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

simplify

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

lo g_{2} (x - 1)

Solution steps

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

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

Simplify

\frac{x ^{2} - 1}{x + 1} : x - 1

= lo g_{2} (x - 1)

lo g_{5} (5 x)

ln (\frac{1}{x - 3})

2 lo g_{10} (a) - 3 lo g_{10} (b) - lo g_{10} (c)

2 ln (5 x + 3) - 5 ln (x - 3) - 4 ln (x)

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