What is 2ln(x)-4ln(x^2+1)+2ln(x-1) ?

The solution to 2ln(x)-4ln(x^2+1)+2ln(x-1) is ln((x^2(x-1)^2)/((x^2+1)^4))

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simplify 2ln(x)-4ln(x^2+1)+2ln(x-1)

simplify

2 ln (x) - 4 ln (x^{2} + 1) + 2 ln (x - 1)

ln (\frac{x ^{2} ( x - 1 ) ^{2}}{( x ^{2} + 1 ) ^{4}})

Solution steps

2 ln (x) - 4 ln (x^{2} + 1) + 2 ln (x - 1)

Apply log rule:

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

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

= ln (x^{2}) - 4 ln (x^{2} + 1) + 2 ln (x - 1)

Apply log rule:

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

4 ln (x^{2} + 1) = ln ((x^{2} + 1)^{4})

= ln (x^{2}) - ln ((x^{2} + 1)^{4}) + 2 ln (x - 1)

Apply log rule:

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

ln (x^{2}) - ln ((x^{2} + 1)^{4}) = ln (\frac{x ^{2}}{( x ^{2} + 1 ) ^{4}})

= ln (\frac{x ^{2}}{( x ^{2} + 1 ) ^{4}}) + 2 ln (x - 1)

Apply log rule:

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

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

= ln (\frac{x ^{2}}{( x ^{2} + 1 ) ^{4}}) + ln ((x - 1)^{2})

Apply log rule:

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

ln (\frac{x ^{2}}{( x ^{2} + 1 ) ^{4}}) + ln ((x - 1)^{2}) = ln (\frac{x ^{2}}{( x ^{2} + 1 ) ^{4}} (x - 1)^{2})

= ln (\frac{x ^{2}}{( x ^{2} + 1 ) ^{4}} (x - 1)^{2})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= ln (\frac{x ^{2} ( x - 1 ) ^{2}}{( x ^{2} + 1 ) ^{4}})

ln^{2} (\frac{- ln ( 2 )}{2})

ln (x^{10} 5 - x)

lo g_{7} (8 yz)

lo g_{11} (\frac{11}{y})

3.13 + lo g_{10} (\frac{[ 0.5 ]}{[ 0.3 ]})

What is 2ln(x)-4ln(x^2+1)+2ln(x-1) ?
The solution to 2ln(x)-4ln(x^2+1)+2ln(x-1) is ln((x^2(x-1)^2)/((x^2+1)^4))