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.
