vereinfachen 24-12ln(1+e^2)+12ln(2)

Lösung

vereinfachen

24 - 12 ln (1 + e^{2}) + 12 ln (2)

24 + ln (\frac{4096}{( 1 + e ^{2} ) ^{12}})

Dezimale

6.79463 \dots

Schritte zur Lösung

24 - 12 ln (1 + e^{2}) + 12 ln (2)

Wende die log Regel an:

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

12 ln (e^{2} + 1) = ln ((e^{2} + 1)^{12})

= 24 - ln ((e^{2} + 1)^{12}) + 12 ln (2)

Wende die log Regel an:

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

12 ln (2) = ln (2^{12})

= 24 - ln ((e^{2} + 1)^{12}) + ln (2^{12})

Wende die log Regel an:

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

ln (2^{12}) - ln ((1 + e^{2})^{12}) = ln (\frac{2 ^{12}}{( e ^{2} + 1 ) ^{12}})

= 24 + ln (\frac{2 ^{12}}{( e ^{2} + 1 ) ^{12}})

2^{12} = 4096

= 24 + ln (\frac{4096}{( 1 + e ^{2} ) ^{12}})

s im pl i f y lo g_{10} (1 0^{2.6}) - 58.56 \cdot lo g_{10} (x)

s im pl i f y ln (x (1 - e^{- \frac{t}{y}}))

e x p an d lo g_{10} (3 m^{4} n^{- 2})

s im pl i f y \frac{- 1}{42} ln (\frac{74}{100})

e x p an d ln ((2 e^{b D} cosh (b u h) + 1)^{N})