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

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

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

simplify

2 ln ∣ 2 x + 1 ∣ - 2 ln ∣ x + 1 ∣

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

Solution steps

2 ln ∣ 2 x + 1 ∣ - 2 ln ∣ x + 1 ∣

Apply log rule:

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

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

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

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

Simplify

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

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

ln (\frac{298}{297})

[ln (s^{2} - 1) - ln (s^{2} + 1)]

5.37 \cdot 1 0^{- 7} + lo g_{10} (\frac{0.0056}{0.0266})

e ln (2 e)

\frac{3}{2} lo g_{10} (15)

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