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Question

Find the value of x from following equation:
logx4+logx16+logx64=12

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Solution

logx4+logx16+logx64=12
logx4+logx42+logx43=12
logx4+2logx4+3logx4=12
6logx4=12
logx4=2
logx22=2
2logx2=2
logx2=1
logx2=log22
On comparing, x=2

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