What is 2ln|l+d|-ln|d|-ln|2l+d| ?

The solution to 2ln|l+d|-ln|d|-ln|2l+d| is ln(((l+d)^2)/(|d||2l+d|))

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simplify 2ln|l+d|-ln|d|-ln|2l+d|

simplify

2 ln ∣ l + d ∣ - ln ∣ d ∣ - ln ∣ 2 l + d ∣

ln (\frac{( l + d ) ^{2}}{∣ d ∣ ∣ 2 l + d ∣})

Solution steps

2 ln ∣ l + d ∣ - ln ∣ d ∣ - ln ∣ 2 l + d ∣

Apply log rule:

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

2 ln ∣ d + l ∣ = ln (∣ d + l ∣^{2})

= ln (∣ d + l ∣^{2}) - ln ∣ d ∣ - ln ∣ d + 2 l ∣

Apply log rule:

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

ln (∣ l + d ∣^{2}) - ln ∣ d ∣ = ln (\frac{∣ d + l ∣ ^{2}}{∣ d ∣})

= ln (\frac{∣ d + l ∣ ^{2}}{∣ d ∣}) - ln ∣ d + 2 l ∣

∣ l + d ∣^{2} = (l + d)^{2}

= ln (\frac{( l + d ) ^{2}}{∣ d ∣}) - ln ∣ 2 l + d ∣

Apply log rule:

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

ln (\frac{( l + d ) ^{2}}{∣ d ∣}) - ln ∣ 2 l + d ∣ = ln \frac{\frac{( d + l ) ^{2}}{∣ d ∣}}{∣ d + 2 l ∣}

= ln \frac{\frac{( l + d ) ^{2}}{∣ d ∣}}{∣ 2 l + d ∣}

Apply the fraction rule:

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

= ln (\frac{( l + d ) ^{2}}{∣ d ∣ ∣ 2 l + d ∣})

ln (\frac{( 3 x + 4 ) ^{3}}{3 ( x ^{2} - 12 ) ^{4}})

lo g_{10} (\frac{1}{5.6 \cdot 1 0 ^{- 2}})

ln (1 - 3) + ln (1 + 3)

lo g_{e} (\frac{1}{x + 1})

\frac{1}{2} ln (28)

What is 2ln|l+d|-ln|d|-ln|2l+d| ?
The solution to 2ln|l+d|-ln|d|-ln|2l+d| is ln(((l+d)^2)/(|d||2l+d|))