Prove that 2log5+log8-1/2log4=2
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Answered by
9
step 1.alogb = log (b^a)
so 2log 5 = log25
1/2log4= log 2
step 2. Log a + log b= Log ab
and log a - log b = log (a/b)
applying we get
log25 + log8 = log (25*8) = log 200
log 200 - log 2 = log (200/2) = log 100
step 3.
we know log 100 can be written as 2 log10 and
log 10= 1
Therefore required LHS =2= RHS
so 2log 5 = log25
1/2log4= log 2
step 2. Log a + log b= Log ab
and log a - log b = log (a/b)
applying we get
log25 + log8 = log (25*8) = log 200
log 200 - log 2 = log (200/2) = log 100
step 3.
we know log 100 can be written as 2 log10 and
log 10= 1
Therefore required LHS =2= RHS
Answered by
4
2log5+log8-1/2log4=2
Log5^2+log8-log√4=2
Log25+log8-log2=2
Log(25×8÷2)=2
Log(100)=2
2=2
Therefore RHS= LHS
Hope it helps you
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