What is 1/2 ln(x+1)+1/2 ln(x-1) ?

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

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simplify 1/2 ln(x+1)+1/2 ln(x-1)

simplify

\frac{1}{2} ln (x + 1) + \frac{1}{2} ln (x - 1)

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

Solution steps

\frac{1}{2} ln (x + 1) + \frac{1}{2} ln (x - 1)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{2} ln (x + 1) = ln ((x + 1)^{\frac{1}{2}})

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{2} ln (x - 1) = ln ((x - 1)^{\frac{1}{2}})

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

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

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

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

\frac{1}{4} ln (\frac{3 tan ( \frac{x}{2} + 1 )}{tan ( \frac{x}{2} + 3 )}) + c

3 (ln (6) - ln (3))

\frac{1}{6} lo g_{2} (\frac{1}{6}) - \frac{5}{6} lo g_{2} (\frac{5}{6})

ln (x) - ln (- x + 1)

ln (+ x y^{2} z^{23} (x x^{2} x^{3} y^{2}) 2)

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