6.0206=-10log_{10}((1-x)^2)

Lösung

6.0206 = - 10 lo g_{10} ((1 - x)^{2})

x = - \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1, x = \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1

Dezimale

x = 0.50000 \dots, x = 1.49999 \dots

Schritte zur Lösung

6.0206 = - 10 lo g_{10} ((1 - x)^{2})

Tausche die Seiten

- 10 lo g_{10} ((1 - x)^{2}) = 6.0206

Teile beide Seiten durch

- 10

lo g_{10} ((1 - x)^{2}) = - \frac{6.0206}{10}

Wende die log Regel an

(1 - x)^{2} = \frac{1}{1 0 ^{\frac{6.0206}{10}}}

Löse

(1 - x)^{2} = \frac{1}{1 0 ^{\frac{6.0206}{10}}} : x = - \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1, x = \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1

x = - \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1, x = \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1

Überprüfe die Lösungen:

x = - \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1

Wahr

, x = \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1

Wahr

Die Lösungen sind

x = - \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1, x = \frac{1}{1 0 ^{\frac{6.0206}{10}}} + 1

lo g_{3} (x) = 2.3

lo g_{125} (1) = lo g_{125} (x + 3)

0 = - 8.314 (298) ln (K)

lo g_{4} (x + 3) = lo g_{4} (4 x - 15)

lo g_{2} (- 5 x + 15) = 4