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Logarithm problem which was not uploaded properly earlier. I am really sorry for that. Q.no.2b and please solve Q.no. 3 also.

Logarithm problem which was not uploaded properly earlier. I am really sorry for that. Q.no.2b and please solve Q.no. 3 also.

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2 Answers

Aditya Gupta
2081 Points
4 years ago
divide both sides by log3x.log4x.log5x and write it as
1= 1/log3x + 1/log4x + 1/log5x
now use the standard formula: 1/logab= logba
so 1= logx3 + logx4 + logx5
or 1= logx(3*4*5)
or logxx= logx(3*4*5)
or x= 3*4*5
or x= 60
kindly approve :)
and ask only one ques per thread
Samyak Jain
333 Points
4 years ago
We know that logb a / logb c  =  logc a.
So in the question, logb (logb N) / logb a  =  loga (logb N).
Now, a^logb (logb N) / logb a  =  a^loga (logb N)  =  logb N      ...[\because a^loga c  =  c].

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