What is 4ln|2x+1|-2ln|x+1| ?

The solution to 4ln|2x+1|-2ln|x+1| is ln(((2x+1)^4)/((x+1)^2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

4 ln ∣ 2 x + 1 ∣ - 2 ln ∣ x + 1 ∣

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

Solution steps

4 ln ∣ 2 x + 1 ∣ - 2 ln ∣ x + 1 ∣

Apply log rule:

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

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

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

Apply log rule:

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

2 ln ∣ x + 1 ∣ = ln (∣ x + 1 ∣^{2})

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

Apply log rule:

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

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

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

Simplify

\frac{∣ 2 x + 1 ∣ ^{4}}{∣ x + 1 ∣ ^{2}} : \frac{( 2 x + 1 ) ^{4}}{( x + 1 ) ^{2}}

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

ln (\frac{273}{300})

ln (\frac{e ^{2}}{e ^{3}}) + 2 ln (3)

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

ln (2022) - ln (2020)

2 ln (a) + 5 ln (b) - 7 ln (c)

What is 4ln|2x+1|-2ln|x+1| ?
The solution to 4ln|2x+1|-2ln|x+1| is ln(((2x+1)^4)/((x+1)^2))