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a,b,c are terms of ap then b-a = c-b 2b = a+c ..........1 log2 , log2x-1 , log2x+3 are in ap so 2log2x+1 = log2+log2x+3 loga + logb =logab now expression is log(2x+1)2 = log[2(2x+3)] (2x+1)2 = 2(2x+3) put 2x = t (t+1)2 = 2(t+3) t = 1,5 2x = 1 or 2x = 5 , x = log5/log2 x = 0 cannot be the solution so x = log5/log2 is the required solution
a,b,c are terms of ap then
b-a = c-b
2b = a+c ..........1
log2 , log2x-1 , log2x+3 are in ap so
2log2x+1 = log2+log2x+3
loga + logb =logab
now expression is
log(2x+1)2 = log[2(2x+3)]
(2x+1)2 = 2(2x+3)
put 2x = t
(t+1)2 = 2(t+3)
t = 1,5
2x = 1 or 2x = 5 , x = log5/log2
x = 0 cannot be the solution so x = log5/log2 is the required solution
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