How do you condense $\frac{1}{3} (2 ln (x + 3) + ln x - ln (x^{2} - 1))$ $1/3(2ln(x+3)+lnx-ln(x^2-1))$ ?

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Answer 1 · 2017-09-21T12:03:59

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

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

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

$= \frac{1}{3} ln (\frac{x {(x + 3)}^{2}}{x^{2} - 1})$ $= 1/3 ln ((x(x+3)^2)/(x^2-1))$ [Ans]

How do you condense 1/3(2ln(x+3)+lnx-ln(x^2-1))13(2ln(x+3)+lnx−ln(x2−1))1/3(2ln(x+3)+lnx-ln(x^2-1))?