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If $\log_{7} \log_{5} (x+5x+x)=0$; find the value of $x$.

Question should have been : If log7log5(x+5x+x)=0 , then find x .

Solving accordingly, log7(7x)=5^{0}

=> log7(7x)=1

=> 7x=7 Hence x=1

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