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

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

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

simplify

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

- lo g_{3} (x - 1)

Solution steps

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

Apply log rule:

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

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

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

Simplify

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

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

Simplify

\frac{1}{x - 1} : (x - 1)^{- 1}

= lo g_{3} ((x - 1)^{- 1})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), assuming x > 0

lo g_{3} ((x - 1)^{- 1}) = - 1 \cdot lo g_{3} (x - 1)

= - 1 \cdot lo g_{3} (x - 1)

Apply rule:

1 \cdot a = a

= - lo g_{3} (x - 1)

lo g_{5} (10 x)

lo g_{10} (\frac{x}{yz})

4 lo g_{b} (x) + 5 lo g_{b} (z)

f (x) = ln (\frac{1}{2 x})

f (x) = lo g_{4} (9 x)

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