log4 (x+1) = 2 + log4 (3x-2) How in the world is the answer 33/47?

Let log be log to base 4
log [ (x + 1) / (3x - 2) ] = 2
(x + 1) / (3x - 2) = 16
x + 1 = 48x - 32
33 = 47x
x = 33/47...Show more

log4( + 1) - log4(3x - 2) = 2
log4[(x + 1) / (3x - 2)] = 2
Antilog.
(x + 1) / (3x - 2)= 4^2 = 16
(x + 1) / (3x - 2) = 16
x + 1 = 16(3x - 2)
x + 1 = 48x - 32
47x = 33
x = 33/47

As required!!!!...Show more

Let L denote log(base 4). Now L(x+1) =2 + L(3x-2)...[1] is to be solved for x. Now [1]-->L[(x+1)/(3x-2)]
= 2. Raising LS & RS to the power 4 gives (x+1)/(3x-2) = 2^4 = 16, ie., x+1 = 16(3x-2) = 48x - 32, ie.,
47x-33 = 0, ie., x = (33/47)....Show more

log4 (x+1) - log4 (3x-2)=2

log4 (x+1)/(3x-2) =2

16=(x+1)/(3x-2)

48x-32=x+1

47x=33

x=33/47...Show more

log₄(x+1) = 2 + log₄(3x-2)
log₄(x+1) - log₄(3x-2) = 2
log₄((x+1)/(3x-2)) = 2

take antilog₄ of both sides
(x+1)/(3x-2) = 4²
x+1 = 16(3x-2)
x+1 = 48x - 32
47x = 33
x = 33/47...Show more