Answer :

ariannaalexander2017

Answer:

log(3x+1)+log12−log(2x−5)=0 , solve for x∈R

log(3x+1)+log(2−1)−log(2x−5)=0

log(3x+1)−log2−log(2x−5)=0

log3x+12−log(2x−5)=0

log3x+122x−5=0

log(3x+12×12x−5)=0

log3x+14x−10=0

e0=3x+14x−10

4x−10=3x+1

x = 11

2005400sblackwell

Answer:

x=11

Step-by-step explanation:

log(3x+1)+log12−log(2x−5)=0 , solve for x∈R

log(3x+1)+log(2−1)−log(2x−5)=0

log(3x+1)−log2−log(2x−5)=0

log3x+12−log(2x−5)=0

log3x+122x−5=0

log(3x+12×12x−5)=0

log3x+14x−10=0

e0=3x+14x−10

4x−10=3x+1

x = 11

Other Questions