What is simplify-4ln|x|-3ln|2x+1|+11ln|x+1| ?

The solution to simplify-4ln|x|-3ln|2x+1|+11ln|x+1| is ln((|x+1|^{11})/(x^4|2x+1|^3))

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simplify-4ln|x|-3ln|2x+1|+11ln|x+1|

simplify

- 4 ln ∣ x ∣ - 3 ln ∣ 2 x + 1 ∣ + 11 ln ∣ x + 1 ∣

ln (\frac{∣ x + 1 ∣ ^{11}}{x ^{4} ∣ 2 x + 1 ∣ ^{3}})

Solution steps

- 4 ln ∣ x ∣ - 3 ln ∣ 2 x + 1 ∣ + 11 ln ∣ x + 1 ∣

Apply log rule:

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

4 ln ∣ x ∣ = ln (∣ x ∣^{4})

= - ln (∣ x ∣^{4}) - 3 ln ∣ 2 x + 1 ∣ + 11 ln ∣ x + 1 ∣

Apply log rule:

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

3 ln ∣ 2 x + 1 ∣ = ln (∣ 2 x + 1 ∣^{3})

= - ln (∣ x ∣^{4}) - ln (∣ 2 x + 1 ∣^{3}) + 11 ln ∣ x + 1 ∣

Rewrite

- ln (∣ x ∣^{4}) - ln (∣ 2 x + 1 ∣^{3})

- (ln (∣ x ∣^{4}) + ln (∣ 2 x + 1 ∣^{3}))

= - (ln (∣ x ∣^{4}) + ln (∣ 2 x + 1 ∣^{3})) + 11 ln ∣ x + 1 ∣

Apply log rule:

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

ln (∣ x ∣^{4}) + ln (∣ 2 x + 1 ∣^{3}) = ln (∣ x ∣^{4} ∣ 2 x + 1 ∣^{3})

= - ln (∣ x ∣^{4} ∣ 2 x + 1 ∣^{3}) + 11 ln ∣ x + 1 ∣

∣ x ∣^{4} = x^{4}

= - ln (x^{4} ∣ 2 x + 1 ∣^{3}) + 11 ln ∣ x + 1 ∣

Apply log rule:

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

11 ln ∣ x + 1 ∣ = ln (∣ x + 1 ∣^{11})

= - ln (x^{4} ∣ 2 x + 1 ∣^{3}) + ln (∣ x + 1 ∣^{11})

Apply log rule:

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

ln (∣ x + 1 ∣^{11}) - ln (x^{4} ∣ 2 x + 1 ∣^{3}) = ln (\frac{∣ x + 1 ∣ ^{11}}{x ^{4} ∣ 2 x + 1 ∣ ^{3}})

= ln (\frac{∣ x + 1 ∣ ^{11}}{x ^{4} ∣ 2 x + 1 ∣ ^{3}})

2 lo g_{2} (x) + 8 lo g_{2} (y)

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

ln (\frac{x ^{4} y ^{2}}{z})

- \frac{ln ( \frac{5}{4} ) - ln ( 7 ) + ln ( 5 )}{2}

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

What is simplify-4ln|x|-3ln|2x+1|+11ln|x+1| ?
The solution to simplify-4ln|x|-3ln|2x+1|+11ln|x+1| is ln((|x+1|^{11})/(x^4|2x+1|^3))