I believe it was Steve who had answered you earlier.
He started by letting u = 3^x
look at the first term 9^(x+2) <--- I will assume those necessary brackets, as did Steve.
9^(x+2) = (3^2)^(x+2)
= 3^(2x + 4)
= 3^2x * 3^4
= 81 * (3^x)^2 , also 6*3^(x+1) = 6*3^x * 3^1 = 18*3^x
so, 9^(x+2) - 6* 3^(x+1) = 0 becomes
81 * (3^x)^2 - 18*3^x = 0
81 u^2 - 18u = 0
9u(9u - 2) = 0
9u = 0 ---> u = 0
or
9u = 2 ---> u = 2/9
so 3^x = 0 , which has no solution
OR
3^x = 2/9
xlog3 = log2 - log9
x = (log2 - log9)/log3 = appr -1.369
This answer can be verified in the original equation
9^x+2 - 6* 3^x+1= 0 find x
Answer earlier I got confused
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