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ln(e^5x)=15
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since \[\Large \ln(a)^b==b \ln(a)\] so it becomes \[\Large \implies 5x(\ln(e))=15\] also since \[\Large \ln(e)=1\] (because the base of ln is e) \[\Large \implies 5x(1)=15\] \[\Large \implies 5x=15\] does this help ?
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