First note that by using the change of base formula
log_9x=(log_3x)/(log_3"9")=(log_3x)/2.
Therefore
2log_9x=log_3x, and
log_3x=log_9x^2.
So we can transform
log_3(sqrt(x)+abs(sqrt(x)-1))=log_9(4sqrt(x)-3+4abs(sqrt(x)-1))
to
log_9(sqrt(x)+abs(sqrt(x)-1))^2=log_9(4sqrt(x)-3+4abs(sqrt(x)-1)).
Since the logarithm is a function, if the logs are equal then their arguments are equal.
(sqrt(x)+abs(sqrt(x)-1))^2=4sqrt(x)-3+4abs(sqrt(x)-1)
Because of the absolute value expressions, there are two cases here - when x>=1 and when 0<=x<1.
When x>=1 we can ignore the absolute value signs and write
(2sqrt(x)-1)^2=4sqrt(x)-3+4(sqrt(x)-1)
4x-4sqrt(x)+1=8sqrt(x)-7
4x-12sqrt(x)+8=0
x-3sqrt(x)+2=0
(sqrt(x)-2)(sqrt(x)-1)=0
sqrt(x)=2 and sqrt(x)=1 which means that x=4, and x=1 are both solutions. In fact, if we plug these values for x back into the original equation, we can see that both of these values work.
For x=1
log_3(sqrt(1)+abs(sqrt(1)-1))=log_9(4sqrt(1)-3+4abs(sqrt(1)-1))
log_3(1)=log_9(1)
0=0.
For x=4
log_3(sqrt(4)+abs(sqrt(4)-1))=log_9(4sqrt(4)-3+4abs(sqrt(4)-1))
log_3(3)=log_9(9)
1=1.
Now for when 0<=x<1, we can rewrite the equation as
(sqrt(x)-sqrt(x)+1))^2=4sqrt(x)-3+4(-sqrt(x)+1)
1=1
Because this is ALWAYS true, there are an infinite number of solutions between and including 0 and 1.