How do you solve #2 log x = log 2 + log (3x - 4)#?

Given #2log x= log 2 + log(3x-4)#

Step 1: Rewrite the equation as a single logarithm on the right hand side, using the sum to product rule, like this

#logx^2 = log(2*(3x-4))#
#logx^2 = log(6x-8)#

Step 2: Transform into the exponential form with the base of 10 like this (or most simple way to put it, if there are log with same base on each side of equation then we can "drop" the log).

#10^(logx^2) = 10^log(6x-8)#
#x^2 = 6x-8#

Now, we can begin to solve the equation by subtracting #6x# and adding 8 to both sides to get

#x^2 -6x + 8 = 0#

This can be solved by using a factoring method like this …

The factors of 8 that add up to -6 are -2 and -4.

#(x-2)(x-4) = 0#
#x-2 = 0 => x= 2 " " or x-4 = 0 => x= 4#

We will need to the check the solutions above to determine whether there are any extraneous solutions (solution that don't work (aka "check out").

Check #x = 2#.

#2log(2)= log 2 + log(3*2-4)#

#2log 2 = 1 log 2 + 1 log 2#
#2 log 2 = 2 log 2#

Therefore #x= 2# is a solution.

Check #x= 4#

#2log(4) = log 2 + log(3*4-4)#

#2 log 4 = log 2 + log 8#
#log 4^2 = log (2*8)#
#log 16 = log 16#

Therefore #x= 4# is also another solution.