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Beliebt Algebra >

vereinfachen ln(0.1^20.5^21^2416)

Lösung

vereinfachen

ln (0. 1^{2} \cdot 0. 5^{2} \cdot 1^{2} \cdot 4 \cdot 16)

Lösung

2 ln (0.1) + 2 ln (0.5) + 6 ln (2)

+1

Dezimale

- 1.83258 \dots

Schritte zur Lösung

ln (0. 1^{2} \cdot 0. 5^{2} \cdot 1^{2} \cdot 4 \cdot 16)

Wende die log Regel an:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (0. 1^{2} \cdot 0. 5^{2} \cdot 1^{2} \cdot 4 \cdot 16) = ln (0. 1^{2}) + ln (0. 5^{2}) + ln (1^{2}) + ln (4) + ln (16)

= ln (0. 1^{2}) + ln (0. 5^{2}) + ln (1^{2}) + ln (4) + ln (16)

Wende die log Regel an:

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

ln (0. 1^{2}) = 2 ln (0.1)

= 2 ln (0.1) + ln (0. 5^{2}) + ln (1^{2}) + ln (4) + ln (16)

Wende die log Regel an:

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

ln (0. 5^{2}) = 2 ln (0.5)

= 2 ln (0.1) + 2 ln (0.5) + ln (1^{2}) + ln (4) + ln (16)

Wende die log Regel an:

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

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

= 2 ln (0.1) + 2 ln (0.5) + 2 ln (1) + ln (4) + ln (16)

2 ln (1) = 0

= 2 ln (0.1) + 2 ln (0.5) + 0 + ln (4) + ln (16)

2 ln (0.1) + 2 ln (0.5) + 0 + ln (4) + ln (16) = 2 ln (0.1) + 2 ln (0.5) + ln (4) + ln (16)

= 2 ln (0.1) + 2 ln (0.5) + ln (4) + ln (16)

ln (4) = 2 ln (2)

ln (16) = 4 ln (2)

= 2 ln (0.1) + 2 ln (0.5) + 2 ln (2) + 4 ln (2)

Addiere gleiche Elemente:

2 ln (2) + 4 ln (2) = 6 ln (2)

= 2 ln (0.1) + 2 ln (0.5) + 6 ln (2)

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