vereinfachen-1/14 ln|3x-1|-3ln|3-2|+32/7 ln|3+3|

Lösung

vereinfachen

- \frac{1}{14} ln ∣ 3 x - 1 ∣ - 3 ln ∣ 3 - 2 ∣ + \frac{32}{7} ln ∣ 3 + 3 ∣

ln (\frac{1296 \cdot 6 ^{\frac{4}{7}} ∣ 3 x - 1 ∣ ^{\frac{13}{14}}}{∣ 3 x - 1 ∣})

Schritte zur Lösung

- \frac{1}{14} ln ∣ 3 x - 1 ∣ - 3 ln ∣ 3 - 2 ∣ + \frac{32}{7} ln ∣ 3 + 3 ∣

Wende die log Regel an:

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

\frac{1}{14} ln ∣ 3 x - 1 ∣ = ln (∣ 3 x - 1 ∣^{\frac{1}{14}})

= - ln (∣ 3 x - 1 ∣^{\frac{1}{14}}) - 3 ln ∣ 3 - 2 ∣ + \frac{32}{7} ln ∣ 3 + 3 ∣

Wende die log Regel an:

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

\frac{32}{7} ln ∣ 3 + 3 ∣ = ln (∣ 3 + 3 ∣^{\frac{32}{7}})

= - ln (∣ 3 x - 1 ∣^{\frac{1}{14}}) - 3 ln ∣ 3 - 2 ∣ + ln (∣ 3 + 3 ∣^{\frac{32}{7}})

Wende die log Regel an:

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

ln (∣ 3 + 3 ∣^{\frac{32}{7}}) - ln (∣ 3 x - 1 ∣^{\frac{1}{14}}) = ln (\frac{∣ 3 + 3 ∣ ^{\frac{32}{7}}}{∣ 3 x - 1 ∣ ^{\frac{1}{14}}})

= ln (\frac{∣ 3 + 3 ∣ ^{\frac{32}{7}}}{∣ 3 x - 1 ∣ ^{\frac{1}{14}}}) - 3 ln ∣ 3 - 2 ∣

Vereinfache

\frac{∣ 3 + 3 ∣ ^{\frac{32}{7}}}{∣ 3 x - 1 ∣ ^{\frac{1}{14}}} : \frac{1296 \cdot 6 ^{\frac{4}{7}} ∣ 3 x - 1 ∣ ^{\frac{13}{14}}}{∣ 3 x - 1 ∣}

= ln (\frac{1296 \cdot 6 ^{\frac{4}{7}} ∣ 3 x - 1 ∣ ^{\frac{13}{14}}}{∣ 3 x - 1 ∣}) - 3 ln ∣ 3 - 2 ∣

3 ln ∣ 3 - 2 ∣ = 0

= ln (\frac{1296 \cdot 6 ^{\frac{4}{7}} ∣ 3 x - 1 ∣ ^{\frac{13}{14}}}{∣ 3 x - 1 ∣}) - 0

ln (\frac{1296 \cdot 6 ^{\frac{4}{7}} ∣ 3 x - 1 ∣ ^{\frac{13}{14}}}{∣ 3 x - 1 ∣}) - 0 = ln (\frac{1296 \cdot 6 ^{\frac{4}{7}} ∣ 3 x - 1 ∣ ^{\frac{13}{14}}}{∣ 3 x - 1 ∣})

= ln (\frac{1296 \cdot 6 ^{\frac{4}{7}} ∣ 3 x - 1 ∣ ^{\frac{13}{14}}}{∣ 3 x - 1 ∣})

j o in \frac{2 ln ( 3 )}{ln ( 7 )} + 6

e x p an d lo g_{10} (\frac{3 x y ^{4}}{z ^{2}})

s im pl i f y ln (\frac{27}{2})

s im pl i f y lo g_{3} (27 b)

s im pl i f y lo g_{10} (\frac{x y ^{2}}{z})