vereinfachen 5^{log_{25}(x+1)}

Lösung

vereinfachen

5^{l o g_{25} (x + 1)}

x + 1

Schritte zur Lösung

5^{l o g_{25} (x + 1)}

Faktorisiere die Zahl:

25 = 5^{2}

= 5^{l o g_{5^{2}} (x + 1)}

Wende die log Regel an:

lo g_{a^{b}} (x) = \frac{1}{b} lo g_{a} (x), a > 0

lo g_{5^{2}} (x + 1) = \frac{1}{2} lo g_{5} (x + 1)

= 5^{\frac{1}{2} l o g_{5} (x + 1)}

Wende Exponentenregel an:

a^{b c} = (a^{b})^{c}, a \geq 0

5^{\frac{1}{2} l o g_{5} (x + 1)} = (5^{l o g_{5} (x + 1)})^{\frac{1}{2}}

= (5^{l o g_{5} (x + 1)})^{\frac{1}{2}}

Wende die log Regel an:

a^{l o g_{a} (b)} = b

5^{l o g_{5} (x + 1)} = x + 1

= (x + 1)^{\frac{1}{2}}

Wende Exponentenregel an:

a^{\frac{1}{2}} = a

(x + 1)^{\frac{1}{2}} = x + 1

= x + 1

s im pl i f y (- 8 x - 12) + (9 x + 4)

f a c t or z^{2} + z - 12

f a c t or x^{4} - 11 x^{2} + 10

s im pl i f y (3 x - 5)^{2} + (5 x + 2)^{2}

f (x) = (x)^{3}