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

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

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

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Multiply fractions:

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

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

ln (2) \cdot \frac{1}{3}

5 ln (3) + ln (4)

\frac{ln ( 2.1 ) - ln ( 2 )}{0.1}

\frac{1}{4} ln (0 - 1) - \frac{1}{4} ln (0 + 3)

ln (z^{4} e^{2 x})

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