SOLUTION: logb(x-1) + logb3 = logbx I thought answer was -1.5 , that seems to be wrong. First I applied the product rule, so log_b(x+1) x log_b(3) = log_b(x) log_b(3x+3) =lob

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Question 1099231: logb(x-1) + logb3 = logbx
I thought answer was -1.5 , that seems to be wrong.
First I applied the product rule, so
log_b(x+1) x log_b(3) = log_b(x)
log_b(3x+3) =lob_b(X)
log_b(3x+3)-log_b(x)=0
Raise all the terms to power of log b
3x+3-x=0
2x+3=0
2x=-3
x=-3/2
Found 2 solutions by Alan3354, Fombitz:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
logb(x-1) + logb3 = logbx
I thought answer was -1.5 , that seems to be wrong.
First I applied the product rule, so
log_b(x+1) x log_b(3) = log_b(x)
log_b(3x+3) =lob_b(X)
log_b(3x+3)-log_b(x)=0
Raise all the terms to power of log b
3x+3-x=0
2x+3=0
2x=-3
x=-3/2
=================
Your method and calculations are correct, but that gives
log(-5/2) + log(3) = log(-3/2)
Logs of negative numbers are not allowed.
--> no solution

Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website!
First line of your solution should be,

SOLUTION: logb(x-1) + logb3 = logbx I thought answer was -1.5 , that seems to be wrong. First I applied the product rule, so log_b(x+1) x log_b(3) = log_b(x) log_b(3x+3) =lob_b(X) lo