vereinfachen-log_{10}(2pi)-1/2 (0.1^2+0.3^2)-2.238

Lösung

vereinfachen

- lo g_{10} (2 π) - \frac{1}{2} (0. 1^{2} + 0. 3^{2}) - 2.238

- lo g_{10} (2) - lo g_{10} (π) - \frac{0.1}{2} - 2.238

Dezimale

- 3.08617 \dots

Schritte zur Lösung

- lo g_{10} (2 π) - \frac{1}{2} (0. 1^{2} + 0. 3^{2}) - 2.238

Wende die log Regel an:

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

lo g_{10} (2 π) = lo g_{10} (2) + lo g_{10} (π)

= - (lo g_{10} (2) + lo g_{10} (π)) - \frac{1}{2} (0. 1^{2} + 0. 3^{2}) - 2.238

\frac{1}{2} (0. 1^{2} + 0. 3^{2}) = 0.1 \cdot \frac{1}{2}

= - (lo g_{10} (2) + lo g_{10} (π)) - 0.1 \cdot \frac{1}{2} - 2.238

\frac{1}{2} \cdot 0.1 = \frac{0.1}{2}

= - (lo g_{10} (2) + lo g_{10} (π)) - \frac{0.1}{2} - 2.238

- (lo g_{10} (2) + lo g_{10} (π)) : - lo g_{10} (2) - lo g_{10} (π)

= - lo g_{10} (2) - lo g_{10} (π) - \frac{0.1}{2} - 2.238

s im pl i f y lo g_{10} (42.61) - lo g_{10} (26.03)

s im pl i f y - lo g_{10} (4.25 \cdot 1 0^{- 7})

s im pl i f y 2 lo g_{2} (x) + lo g_{2} (y)

s im pl i f y \frac{1}{32} ln (x + 2) - \frac{1}{32} ln (x - 2)

e x p an d lo g_{9} ((\frac{8 ^{5}}{11})^{4})