What is 3[ln(x-2)+2ln(x+1)-ln(x+2)-5ln(x-1)] ?

The solution to 3[ln(x-2)+2ln(x+1)-ln(x+2)-5ln(x-1)] is 3ln(((x-2)(x+1)^2)/((x+2)(x-1)^5))

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simplify 3[ln(x-2)+2ln(x+1)-ln(x+2)-5ln(x-1)]

simplify

3 [ln (x - 2) + 2 ln (x + 1) - ln (x + 2) - 5 ln (x - 1)]

3 ln (\frac{( x - 2 ) ( x + 1 ) ^{2}}{( x + 2 ) ( x - 1 ) ^{5}})

Solution steps

3 [ln (x - 2) + 2 ln (x + 1) - ln (x + 2) - 5 ln (x - 1)]

Apply log rule:

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

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

= 3 [ln (x - 2) + ln ((x + 1)^{2}) - ln (x + 2) - 5 ln (x - 1)]

Apply log rule:

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

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

= 3 [ln ((x - 2) (x + 1)^{2}) - ln (x + 2) - 5 ln (x - 1)]

Apply log rule:

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

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

= 3 [ln (\frac{( x - 2 ) ( x + 1 ) ^{2}}{x + 2}) - 5 ln (x - 1)]

Apply log rule:

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

5 ln (x - 1) = ln ((x - 1)^{5})

= 3 [ln (\frac{( x - 2 ) ( x + 1 ) ^{2}}{x + 2}) - ln ((x - 1)^{5})]

Apply log rule:

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

ln (\frac{( x - 2 ) ( x + 1 ) ^{2}}{x + 2}) - ln ((x - 1)^{5}) = ln (\frac{\frac{( x + 1 ) ^{2} ( x - 2 )}{x + 2}}{( x - 1 ) ^{5}})

= 3 [ln (\frac{\frac{( x - 2 ) ( x + 1 ) ^{2}}{x + 2}}{( x - 1 ) ^{5}})]

Apply the fraction rule:

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

= 3 [ln (\frac{( x - 2 ) ( x + 1 ) ^{2}}{( x + 2 ) ( x - 1 ) ^{5}})]

Remove parentheses:

(a) = a

= 3 ln (\frac{( x - 2 ) ( x + 1 ) ^{2}}{( x + 2 ) ( x - 1 ) ^{5}})

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What is 3[ln(x-2)+2ln(x+1)-ln(x+2)-5ln(x-1)] ?
The solution to 3[ln(x-2)+2ln(x+1)-ln(x+2)-5ln(x-1)] is 3ln(((x-2)(x+1)^2)/((x+2)(x-1)^5))