find the solution for:
log(3x-1)+log2=log4+log(x+2)

Question

find the solution for: 
log(3x-1)+log2=log4+log(x+2)

slaaibak · Answer

Use this:

log a + log b = log ab

rulnick · Answer

log(3x-1)-log(x+2) = log4 - log2
log((3x-1)/(x+2)) = log(4/2)
(3x-1)/(x+2) = 2
3x-1=2x+4
x=5

campbell_st · Answer

simplify the logs 1st

log(3x-1) + log 2 = log(2(3x -1) or  log(6x - 2)

log 4 + log( x + 2) = log(4(x+2) = log(4x + 8)

now equate both sides and solve for x

6x - 2 = 4x + 8

2x = 10

x = 5

anonymous · Answer

Thanks:) would you solve log(4x-1)=log(x+1)+log2 the same way

campbell_st · Answer

log(x+1) + log 2 = log(2(x+1) or log(2x + 2)

so log(4x -1) = log(2x +2)
  equate and solve for x

4x -1 = 2x + 2

2x = 3

x = 3/2