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

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

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

simplify

2 (2 ln ∣ x + 2 ∣ - ln ∣ x + 1 ∣)

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

Solution steps

2 (2 ln ∣ x + 2 ∣ - ln ∣ x + 1 ∣)

Apply log rule:

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

2 ln ∣ x + 2 ∣ = ln (∣ x + 2 ∣^{2})

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

Apply log rule:

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

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

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

∣ x + 2 ∣^{2} = (x + 2)^{2}

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

- \frac{1}{0.024} ln (\frac{6 - p}{4.2})

\frac{19 ln ( \frac{y}{x} + 5 ) - ln ( \frac{y}{x} )}{5} - \frac{y}{x}

ln (\frac{x}{\frac{x}{2}})

ln (\frac{20474}{375}) \cdot 4

10 lo g_{10} (\frac{\frac{2}{π}}{1 0 ^{- 12}})

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