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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Simplify

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

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

ln (a + b) + 5 ln (a - b) - 4 ln (c)

lo g_{9} (v) + 6 lo g_{9} (w) + \frac{1}{2} lo g_{9} (u)

lo g_{2} (5 n)

lo g_{10} (2 a) + lo g_{10} (b) - 2 lo g_{10} (ab)

\frac{1}{3} lo g_{c} (x) + 4 lo g_{c} (y) - 2 lo g_{c} (x)

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