pls someone tell fast
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hi friend,
the. correct option is →D
which is a=c^b/c
now given ,
=>10^(log base a(log base b(log x base c))=1
applying logarithm on both sides, we get
=>log 1 base 10=(log base a(log base b(log x base c))
=>0=(log base a(log base b(log x base c)
so,to satisfy this condition,
=>(log base b(log x base c) =1
converting into exponential form
=>b=logx base c
again converting into exponential for
=>c^b=x-------(1)
and also given 10^(log base b(log base c(logx base a))=1
applying logarithm on both sides, we get
=>log 1 base 10=log base b(log base c(logx base a))
=>0=log base b(log base c(logx base a))
so,to satisfy this condition,
=>(log base c(logx base a)) =1
converting into exponential form
=>logx base a=c
again,
converting into exponential form
we get
=>x=a^c
by (1)
=>a^c=c^b
=>a=c^(b/c)
I think it is little confusing...but still I hope it will help u :)
the. correct option is →D
which is a=c^b/c
now given ,
=>10^(log base a(log base b(log x base c))=1
applying logarithm on both sides, we get
=>log 1 base 10=(log base a(log base b(log x base c))
=>0=(log base a(log base b(log x base c)
so,to satisfy this condition,
=>(log base b(log x base c) =1
converting into exponential form
=>b=logx base c
again converting into exponential for
=>c^b=x-------(1)
and also given 10^(log base b(log base c(logx base a))=1
applying logarithm on both sides, we get
=>log 1 base 10=log base b(log base c(logx base a))
=>0=log base b(log base c(logx base a))
so,to satisfy this condition,
=>(log base c(logx base a)) =1
converting into exponential form
=>logx base a=c
again,
converting into exponential form
we get
=>x=a^c
by (1)
=>a^c=c^b
=>a=c^(b/c)
I think it is little confusing...but still I hope it will help u :)
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