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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

lo g_{2} (x^{2}) - 3 lo g_{2} (x + 1) + lo g_{4} (3 x)

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

\frac{3}{4} lo g_{2} (\frac{3}{4}) + \frac{1}{4} lo g_{2} (\frac{1}{4})

ln (\frac{x ^{4} y}{z ^{5}})

lo g_{4} ((x y)^{3}) - lo g_{4} (x y)

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