log_{7}(3^{2x}-3x+1)=1

Lösung

lo g_{7} (3^{2 x} - 3 x + 1) = 1

x = 1, x = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{2 ln ( 3 )}

Dezimale

x = 1, x = - 1.99584 \dots

Schritte zur Lösung

lo g_{7} (3^{2 x} - 3 x + 1) = 1

Wende die log Regel an

9^{x} - 3 x + 1 = 7

Löse

9^{x} - 3 x + 1 = 7 : x = 1, x = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{2 ln ( 3 )}

x = 1, x = - \frac{W _{0} ( - \frac{2 l n ( 3 )}{243} ) + 4 ln ( 3 )}{2 ln ( 3 )}

lo g_{49} (x - 6) = 0.5

2 = lo g_{x} (9)

ln (x^{2} - 63) = 0

x - 3 ln (x) = 0

lo g_{4} ((n + 3)^{4}) = 4