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

The solution to 2ln(2+1)+ln(2+1)+1-2-2ln(2) is ln(27/4)-1

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

simplify

2 ln (2 + 1) + ln (2 + 1) + 1 - 2 - 2 ln (2)

ln (\frac{27}{4}) - 1

Decimal Notation

0.90954 \dots

Solution steps

2 ln (2 + 1) + ln (2 + 1) + 1 - 2 - 2 ln (2)

Apply log rule:

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

2 ln (2) = ln (2^{2})

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

Apply log rule:

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

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

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

\frac{2 + 1}{2 ^{2}} = \frac{3}{4}

= 2 ln (2 + 1) + ln (\frac{3}{4}) + 1 - 2

Apply log rule:

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

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

= ln ((2 + 1)^{2}) + ln (\frac{3}{4}) + 1 - 2

Apply log rule:

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

ln ((2 + 1)^{2}) + ln (\frac{3}{4}) = ln (\frac{3}{4} (2 + 1)^{2})

= ln (\frac{3}{4} (2 + 1)^{2}) + 1 - 2

(2 + 1)^{2} \frac{3}{4} = \frac{27}{4}

= ln (\frac{27}{4}) + 1 - 2

Add/Subtract the numbers:

1 - 2 = - 1

= ln (\frac{27}{4}) - 1

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What is 2ln(2+1)+ln(2+1)+1-2-2ln(2) ?
The solution to 2ln(2+1)+ln(2+1)+1-2-2ln(2) is ln(27/4)-1