solvefor x,log_{e}(x+1)=y+C

Lösung

löse nach

x, lo g_{e} (x + 1) = y + C

x = e^{y + C} - 1

Schritte zur Lösung

lo g_{e} (x + 1) = y + C

Wende die log Regel an

x + 1 = e^{y + C}

Löse

x + 1 = e^{y + C} : x = e^{y + C} - 1

x = e^{y + C} - 1

ln (a) = 722

ln (T) = 1.56 + 0.82 \cdot ln (11)

2 = lo g_{10} (n)

ln (x^{2}) - 7 = 0

2 lo g_{10} (1 + x) = 2