vereinfachen 1/2 ln(5)-1/2 ln(2)

Lösung

vereinfachen

\frac{1}{2} ln (5) - \frac{1}{2} ln (2)

\frac{1}{2} ln (\frac{5}{2})

Dezimale

0.45814 \dots

Schritte zur Lösung

\frac{1}{2} ln (5) - \frac{1}{2} ln (2)

Wende die log Regel an:

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

\frac{1}{2} ln (5) = ln (5^{\frac{1}{2}})

= ln (5^{\frac{1}{2}}) - \frac{1}{2} ln (2)

Wende die log Regel an:

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

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

= ln (5^{\frac{1}{2}}) - ln (2^{\frac{1}{2}})

Wende die log Regel an:

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

ln (5^{\frac{1}{2}}) - ln (2^{\frac{1}{2}}) = ln (\frac{5 ^{\frac{1}{2}}}{2 ^{\frac{1}{2}}})

= ln (\frac{5 ^{\frac{1}{2}}}{2 ^{\frac{1}{2}}})

Fasse gleiche Potenzen zusammen:

\frac{x ^{\frac{a}{m}}}{y ^{\frac{b}{m}}} = m \frac{x ^{a}}{y ^{b}}

= ln (\frac{5}{2})

Schreibe um

= ln ((\frac{5}{2})^{\frac{1}{2}})

Wende log Regel an

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

angenommen

x \geq 0

= \frac{1}{2} ln (\frac{5}{2})

s im pl i f y ln (5 e^{- 4 t})

s im pl i f y \frac{ln ( 16 ) - ln ( 10 )}{ln ( 8 )} \cdot ln (10)

e x p an d lo g_{6} ((\frac{12}{7})^{5})^{5}

e x p an d lo g_{10} (\frac{x ^{6} y ^{18}}{z ^{14}})

s im pl i f y - 50 ln (\frac{13}{16})