What is e^{2(ln(x)-ln(1-x))} ?

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

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simplify e^{2(ln(x)-ln(1-x))}

simplify

e^{2 (l n (x) - l n (1 - x))}

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

Solution steps

e^{2 (l n (x) - l n (1 - x))}

Apply log rule:

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

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

= e^{2 l n (\frac{x}{1 - x})}

Apply exponent rule:

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

= (e^{l n (\frac{x}{1 - x})})^{2}

Apply log rule:

a^{l o g_{a} (b)} = b

e^{l n (\frac{x}{1 - x})} = \frac{x}{1 - x}

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

ln (\frac{86 - 60}{86 - 50})

ln (x^{4} (1 - x)^{14})

ln (\frac{5 e ^{4 x + 3}}{3 x + 2})

- lo g_{10} (3.6 \cdot 1 0^{- 11})

ln (\frac{a}{b ^{- 3}})

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