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

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

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

simplify

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

ln (\frac{x - 2}{x + 2})

Solution steps

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

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

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

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

Apply log rule:

lo g_{e} (x) = ln (x)

= ln (\frac{x - 2}{x + 2})

ln (\frac{1}{cos ( 2 x )})

- ln (\frac{31}{9})

(ln (52) - ln (65)) (\frac{1}{3})

4 (ln (x) - 2 ln (2))

lo g_{2} (5) \cdot \frac{7}{8}

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