log(3^x-8)=3^(2-x).please solve it Share with your friends Share 0 Neha Sethi answered this Dear student There is a mistake in your question.Correct question islog33x-8=2-x⇒log3x-8log3=2-x⇒log3x-8=2-xlog3⇒log3x-8=log32-x⇒3x-8=32-x⇒3x-8=323x⇒3x-8=93xLet 3x=u⇒u-8=9u⇒u2-8u-9=0⇒u2-9u+u-9=0⇒u(u-9)+1(u-9)=0⇒(u-9)(u+1)=0⇒u-9=0 or u+1=0⇒u=9 or u=-1⇒3x=9 or 3x=-1 Rejected⇒3x=32 ⇒x=2Note:1) logalogb=logba2) alogbc=logbca Regards 0 View Full Answer
