What is 2log_{2}(x+1)-log_{2}(x+3)-log_{2}(x-1) ?

The solution to 2log_{2}(x+1)-log_{2}(x+3)-log_{2}(x-1) is log_{2}(((x+1)^2)/((x+3)(x-1)))

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simplify 2log_{2}(x+1)-log_{2}(x+3)-log_{2}(x-1)

simplify

2 lo g_{2} (x + 1) - lo g_{2} (x + 3) - lo g_{2} (x - 1)

lo g_{2} (\frac{( x + 1 ) ^{2}}{( x + 3 ) ( x - 1 )})

Solution steps

2 lo g_{2} (x + 1) - lo g_{2} (x + 3) - lo g_{2} (x - 1)

Apply log rule:

b \cdot lo g_{a} (x) = lo g_{a} (x^{b})

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

= lo g_{2} ((x + 1)^{2}) - lo g_{2} (x + 3) - 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)^{2}) - lo g_{2} (x + 3) = lo g_{2} (\frac{( x + 1 ) ^{2}}{x + 3})

= lo g_{2} (\frac{( x + 1 ) ^{2}}{x + 3}) - 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} (\frac{( x + 1 ) ^{2}}{x + 3}) - lo g_{2} (x - 1) = lo g_{2} (\frac{\frac{( x + 1 ) ^{2}}{x + 3}}{x - 1})

= lo g_{2} (\frac{\frac{( x + 1 ) ^{2}}{x + 3}}{x - 1})

Simplify

\frac{\frac{( x + 1 ) ^{2}}{x + 3}}{x - 1} : \frac{( x + 1 ) ^{2}}{( x + 3 ) ( x - 1 )}

= lo g_{2} (\frac{( x + 1 ) ^{2}}{( x + 3 ) ( x - 1 )})

\frac{1}{3} ln (2)

ln (a \cdot b)

- lo g_{10} (1.3 \cdot 1 0^{- 4})

5 ln (x) - 4 ln (y)

5 ln (x) - 6 ln (y)

What is 2log_{2}(x+1)-log_{2}(x+3)-log_{2}(x-1) ?
The solution to 2log_{2}(x+1)-log_{2}(x+3)-log_{2}(x-1) is log_{2}(((x+1)^2)/((x+3)(x-1)))