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logbase4(x-1)=logbase2(x-3) have how many sloutions
logbase4(x-1)=logbase2(x-3)
x-3=(x-1)^2
1/2log(x-1) = log(x-3) " here now both sides of equations has same base that is 2"
log (x-1)1/2 =log (x-3) "now we can eliminate the log on both side because they have same base "
(x-1)1/2 =(x-3)
x-1= (x-3)2
x2-7x+10=0
therefore x=2,5
THIS EQUATION WILL HAVE 2 REAL AND DISTINCT SOLUTIONS
log4(x-1)= log2(x-3)
ON SOLVING [log(x-1)/log4] = [ log(x-3)/log2]
[log(x-1)/2log2] = [log(x-3)/log2]
CANCELLING LOG2 BOTH SIDES WE GET
log(x-1)= 2 log (x-3)
log(x-1) = log(x-3)2
TAKING ANTILOGS BOTH SIDES WE GET
(x-1)=(x-3)2
ON SOLVING THE QUADRATIC IN x YOU WILL GET TWO REAL AND DISTINCT VALUES WHICH IS 5,2
HOPE THIS HELPS
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