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

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!!!!

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).

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

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

log₄(x+1) = 2 + log₄(3x−2)

log₄(x+1) − log₄(3x−2) = 2

log₄((x+1)/(3x−2)) = 2

(x+1)/(3x−2) = 4^2

x + 1 = 16 (3x−2)

x + 1 = 48x − 32

33 = 47x

x = 33/47

Log4 (x+1) = 2 + log4 (3x-2)

Log4 (x+1) = 2log4(4) + log4 (3x-2)

Log4 (x+1) = log4(16) + log4 (3x-2

Log4 (x+1) = log4( (3x-2)16)

x+1 = 16(3x-2)

x+1 = 48x - 32

1+32 = 48x-x

x = 33/47

Taking the antilog,

.. x+1 = 4^2*(3x-2)

.. 33 = 47x ... add 32-x

.. 33/47 = x ... divide by 47