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العربية

vereinfachen [3ln(3)-1-2ln(2)]-[3ln(3)-1-2ln(2)]

Lösung

vereinfachen

[3 ln (3) - 1 - 2 ln (2)] - [3 ln (3) - 1 - 2 ln (2)]

0

Schritte zur Lösung

[3 ln (3) - 1 - 2 ln (2)] - [3 ln (3) - 1 - 2 ln (2)]

Wende die log Regel an:

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

3 ln (3) = ln (3^{3})

= [ln (3^{3}) - 1 - 2 ln (2)] - [3 ln (3) - 1 - 2 ln (2)]

Wende die log Regel an:

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

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

= [ln (3^{3}) - 1 - ln (2^{2})] - [3 ln (3) - 1 - 2 ln (2)]

Wende die log Regel an:

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

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

= [ln (\frac{3 ^{3}}{2 ^{2}}) - 1] - [3 ln (3) - 1 - 2 ln (2)]

\frac{3 ^{3}}{2 ^{2}} = \frac{27}{4}

= [ln (\frac{27}{4}) - 1] - [3 ln (3) - 1 - 2 ln (2)]

Wende die log Regel an:

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

3 ln (3) = ln (3^{3})

= [ln (\frac{27}{4}) - 1] - [ln (3^{3}) - 1 - 2 ln (2)]

Wende die log Regel an:

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

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

= [ln (\frac{27}{4}) - 1] - [ln (3^{3}) - 1 - ln (2^{2})]

Wende die log Regel an:

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

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

= [ln (\frac{27}{4}) - 1] - [ln (\frac{3 ^{3}}{2 ^{2}}) - 1]

\frac{3 ^{3}}{2 ^{2}} = \frac{27}{4}

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

Entferne die Klammern:

(a) = a

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

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

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

Vereinfache

ln (\frac{27}{4}) - 1 - ln (\frac{27}{4}) + 1 : 0

= 0

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s im pl i f y lo g_{8} (4 x)

s im pl i f y ln (x y) + ln (\frac{x}{e})

e x p an d ln ((sin (x))^{2})