ln(x+1)=log_{2}(3)

Lösung

ln (x + 1) = lo g_{2} (3)

x = e^{l o g_{2} (3)} - 1

Dezimale

x = 3.87910 \dots

Schritte zur Lösung

ln (x + 1) = lo g_{2} (3)

Wende die log Regel an

x + 1 = e^{l o g_{2} (3)}

Löse

x + 1 = e^{l o g_{2} (3)} : x = e^{l o g_{2} (3)} - 1

x = e^{l o g_{2} (3)} - 1

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

6 ln (4 x - 10) - 2 = 40

ln (x^{4}) = 24

ln (x + 1) = 2 - ln (x - 1)

lo g_{2} (x + a) = b