Aufgabe 261
Schreibe die Quotienten als einzelne Logarithmen an
\(\eqalign{ & \ln \frac{{abc}}{d} = \cr & \ln \frac{d}{{abc}} = \cr & \cr & \ln \left( {a \cdot {b^2} \cdot {c^3}} \right) = \cr & \ln \left( {\frac{{a \cdot {b^2}}}{{{c^3}}}} \right) = \cr & \cr & \ln \left( {\frac{{3 \cdot {a^2}}}{{{b^2} \cdot c}}} \right) = \cr & \cr & \ln \left( {a \cdot \root 4 \of b } \right) = \cr & \ln \left( {\root 4 \of {\frac{{{a^2}}}{b}} } \right) = \cr} \)
Lösungsweg
Unter Verwendung der folgenden Rechenregeln für Logarithmen
\(\eqalign{ & {\log _a}\left( {u \cdot v} \right) = {\log _a}\left( u \right) + {\log _a}\left( v \right) \cr & {\log _a}\left( {\frac{u}{v}} \right) = {\log _a}\left( u \right) - {\log _a}\left( v \right) \cr} \)
können wir wie folgt anschreiben:
\(\eqalign{ & \ln \dfrac{{abc}}{d} = \ln \left( a \right) + \ln \left( b \right) + \ln \left( d \right) - \ln \left( d \right) \cr & \ln \dfrac{d}{{abc}} = \ln \left( d \right) - \left( {\ln \left( a \right) + \ln \left( b \right) + \ln \left( d \right)} \right) = \ln \left( d \right) - \ln \left( a \right) - \ln \left( b \right) - \ln \left( d \right) \cr & \cr & \ln \left( {a \cdot {b^2} \cdot {c^3}} \right) = \ln \left( a \right) + \ln \left( {{b^2}} \right) + \ln \left( {{c^2}} \right) = \ln \left( a \right) + 2 \cdot \ln \left( b \right) + 3 \cdot \ln \left( c \right) \cr & \ln \left( {\dfrac{{a \cdot {b^2}}}{{{c^3}}}} \right) = \ln \left( a \right) + \ln \left( {{b^2}} \right) - \ln \left( {{c^3}} \right) = \ln \left( a \right) + 2 \cdot \ln \left( b \right) - 3 \cdot \ln \left( c \right) \cr & \cr & \ln \left( {\frac{{3 \cdot {a^2}}}{{{b^2} \cdot c}}} \right) = \ln \left( 3 \right) + \ln \left( {{a^2}} \right) - \left[ {\ln \left( {{b^2}} \right) + \ln \left( c \right)} \right] = \ln \left( 3 \right) + 2 \cdot \ln \left( a \right) - 2 \cdot \ln \left( b \right) - \ln \left( c \right) \cr & \ln \left( {a \cdot \root 4 \of b } \right) = \ln \left( {a \cdot {b^{\frac{1}{4}}}} \right) = \ln \left( a \right) + \ln \left( {{b^{\frac{1}{4}}}} \right) = \ln \left( a \right) + \dfrac{1}{4} \cdot \ln \left( b \right) \cr & \ln \left( {\root 4 \of {\dfrac{{{a^2}}}{b}} } \right) = \ln {\left( {\dfrac{{{a^2}}}{b}} \right)^{\dfrac{1}{4}}} = \dfrac{1}{4} \cdot \ln \left( {\dfrac{{{a^2}}}{b}} \right) = \dfrac{1}{4} \cdot \left[ {\ln \left( {{a^2}} \right) - \ln \left( b \right)} \right] = \dfrac{1}{4}\left[ {2 \cdot \ln \left( a \right) - \ln \left( b \right)} \right] \cr} \)
Ergebnis
Die richtige Lösung lautet:
\(\eqalign{ & \ln \dfrac{{abc}}{d} = \ln \left( a \right) + \ln \left( b \right) + \ln \left( d \right) - \ln \left( d \right) \cr & \ln \dfrac{d}{{abc}} = \ln \left( d \right) - \ln \left( a \right) - \ln \left( b \right) - \ln \left( d \right) \cr & \cr & \ln \left( {a \cdot {b^2} \cdot {c^3}} \right) = \ln \left( a \right) + 2 \cdot \ln \left( b \right) + 3 \cdot \ln \left( c \right) \cr & \ln \left( {\dfrac{{a \cdot {b^2}}}{{{c^3}}}} \right) = \ln \left( a \right) + 2 \cdot \ln \left( b \right) - 3 \cdot \ln \left( c \right) \cr & \cr & \ln \left( {\dfrac{{3 \cdot {a^2}}}{{{b^2} \cdot c}}} \right) = \ln \left( 3 \right) + 2 \cdot \ln \left( a \right) - 2 \cdot \ln \left( b \right) - \ln \left( c \right) \cr & \ln \left( {a \cdot \root 4 \of b } \right) = \ln \left( a \right) + \frac{1}{4} \cdot \ln \left( b \right) \cr & \ln \left( {\root 4 \of {\dfrac{{{a^2}}}{b}} } \right) = \dfrac{1}{4}\left[ {2 \cdot \ln \left( a \right) - \ln \left( b \right)} \right] \cr} \)