Question

324+3 - 24303 +9=0. Find a.

JEE/Engineering Exams

Maths

Solution

114

4.0 (1 ratings)

( 3^{2 x+3}-243 cdot 3^{x}+9=0 ) ( 9left(3 times 3^{2 x}-27 times 3^{x}+1right)=0 ) ( operatorname{let} 3^{x}=t ) ( 3 t^{2}-27 t+1 ) ( t=frac{27 pm sqrt{9729-12}}{6} ) ( x=log left(frac{27 pm sqrt{717}}{6}right) )