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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

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

= 2 ln (1 - x) - ln (1 + x)

lo g_{10} (x^{2} z)

ln (\frac{75}{27})

ln (\frac{0.16}{12})

ln (6) - ln (4) - ln (7) + ln (5)

ln (x^{2} + x) + I n e - ln (x + 1)

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