What is 3/2 [ln(x^2-1)-ln(x+1)+ln(x-1)] ?

The solution to 3/2 [ln(x^2-1)-ln(x+1)+ln(x-1)] is 3ln(x-1)

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simplify 3/2 [ln(x^2-1)-ln(x+1)+ln(x-1)]

simplify

\frac{3}{2} [ln (x^{2} - 1) - ln (x + 1) + ln (x - 1)]

3 ln (x - 1)

Solution steps

\frac{3}{2} [ln (x^{2} - 1) - ln (x + 1) + ln (x - 1)]

Apply log rule:

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

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

= \frac{3}{2} [ln (\frac{x ^{2} - 1}{x + 1}) + ln (x - 1)]

Simplify

\frac{x ^{2} - 1}{x + 1} : x - 1

= \frac{3}{2} [ln (x - 1) + ln (x - 1)]

Multiply fractions:

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

= \frac{3 ( ln ( x - 1 ) + ln ( x - 1 ) )}{2}

3 (ln (x - 1) + ln (x - 1)) = 6 ln (x - 1)

= \frac{6 ln ( x - 1 )}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 ln (x - 1)

ln (x \frac{y}{z})

lo g_{10} ((\frac{6}{5})^{6})

lo g_{6} (14) - \frac{1}{2} lo g_{6} (49) + lo g_{6} (18)

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

lo g_{5} (x) + \frac{1}{3} lo g_{5} (y) - 4 lo g_{5} (z)

What is 3/2 [ln(x^2-1)-ln(x+1)+ln(x-1)] ?
The solution to 3/2 [ln(x^2-1)-ln(x+1)+ln(x-1)] is 3ln(x-1)