What is ln(x^2+1)+1/2 ln(x+3)-ln(x-1) ?

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

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simplify ln(x^2+1)+1/2 ln(x+3)-ln(x-1)

simplify

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

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

Solution steps

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

Apply log rule:

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

\frac{1}{2} ln (x + 3) = ln ((x + 3)^{\frac{1}{2}})

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

Apply log rule:

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

ln (x^{2} + 1) + ln ((x + 3)^{\frac{1}{2}}) = ln ((x^{2} + 1) (x + 3)^{\frac{1}{2}})

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

Apply log rule:

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

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

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

- ln (i - n) + ln (i)

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

ln (t) + ln (t + 6) + ln (t + 4)

\frac{8 \cdot 12 5 ^{- x}}{3 ln ( 5 )} + C

lo g_{2} (((x - 5)^{3})^{\frac{1}{2}})

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