Wie löse ich diese Logarithmusaufgabe? log_3 (4x - 2) = 1 + log_3 (x+7)

Question

log₃ (4x - 2) = 1 + log₃ (x+7)

x=?

lösung ist 23. 

Comments

log sei  log_3

log(4x-2)-log(x+7) = 1

log((4x-2)/(x+7)) = 1 |3^x

(4x-2)/(x+7) = 3^1

4x-2 = 3x+21

...

Answer 1

Hi,

mein Vorschlag:

 

log₃ (4x - 2) = 1 + log₃ (x+7)               |1 = log₃ (3)

log₃ (4x - 2) = log₃ (3) + log₃ (x+7)   |log-Gesetze

log₃ (4x - 2) = log₃ (3(x+7))                |3 anwenden

4x-2 = 3x+21                                         |-3x+2

x = 23

 

Fertich is die Sach ;).

 

Grüße