2log_{10}(x+1)=x-1

Lösung

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

x = - \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1, x = - \frac{2 W _{0} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1

Dezimale

x = 1.93530 \dots, x = - 0.88597 \dots

Schritte zur Lösung

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

Teile beide Seiten durch

2

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

Wenn

f (x) = g (x),

dann

a^{f (x)} = a^{g (x)}

1 0^{l o g_{10} (x + 1)} = 1 0^{\frac{x - 1}{2}}

Vereinfache

1 0^{l o g_{10} (x + 1)} : x + 1

x + 1 = 1 0^{\frac{x - 1}{2}}

Löse

x + 1 = 1 0^{\frac{x - 1}{2}} : x = - \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1, x = - \frac{2 W _{0} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1

x = - \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1, x = - \frac{2 W _{0} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1

Überprüfe die Lösungen:

x = - \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1

Wahr

, x = - \frac{2 W _{0} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1

Wahr

Die Lösungen sind

x = - \frac{2 W _{- 1} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1, x = - \frac{2 W _{0} ( - \frac{l n ( 10 )}{20} )}{ln ( 10 )} - 1

lo g_{2} (5 + 2 x) - lo g_{2} (4 - x) = 3

lo g_{4} (\frac{3 - x}{7}) = 1

8 lo g_{9} (2) = lo g_{9} (x)

- lo g_{10} (x) = 4.5

lo g_{7} (X - 2) + lo g_{7} (X + 2) = 1