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

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

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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{∣ d ∣ ∣ 2 l + d ∣}{( l + d ) ^{2}})

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

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

∣ l + d ∣^{2} = (l + d)^{2}

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

Apply log rule:

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

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

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

Multiply fractions:

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

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

lo g_{4} (x^{2} - 1) - 5 lo g_{4} (x - 1)

ln (1) - ln (\frac{1}{3})

\frac{10 ln ( 7.01 ) - 10 ln ( 7 )}{7.01 - 7}

\frac{1}{3} \cdot ln (\frac{7}{6} + \frac{5}{2})

\frac{1}{2 105} - ln (21) + ln (20)

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