Question

9∧(x+2) + 6.3∧(x+1) + 1 = 0

Answer 1

9^(x + 2) +6.3^(x +1) +1 =0

use concept ,
x^(m+ n) = x^m.x^n here

so,

9^x.9^2 +6.3^x.3^1 + 1=0

81.9^x + 18.3^x +1 =0

81.(3^2)^x + 18.3^x +1 =0

81.3^2x + 18.3^x +1 =0

let 3^x = r
then 3^2x = r²

so ,

81r² +18r +1 =0

(9r)² +2.(9r).1 + 1 =0

(9r +1 )² =0

r = -1/9

but r = 3^x

so, 3^x = -1/9
but we know, exponential term is always positive .
so, here no solution