Question

how to continue?
lg (3^x-3) -lg2=lg (3^x+9)-lg(3^x-3)

3^x-3/2=3^x+9/3^x-3

(3^x-3)^2 =2(3^x+9)
3^x=a

a^2-6a+9=2a+18
a^2-8a-9=0
D=100
(sqr)D=10
x1=-2 x2=8
3^x=-2
3^x=8
how to continue solving? or is there any mistakes ?
need to find x that will be correct to inequality

Answer 1

i do see one error.
a^2 -8a -9 =0
a^2-9a +a -9 = 0
a(a-9) + 1 (a-9) = 0
(a+1)(a-9) = 0
a=-1,9
3^x = -1,9

lets continue,
3^x cannot be negative as the least value is 0.
so discarding 3^x = -1

3^x = 9
we know 3^2 = 9
x=2

let me know if you have any more doubts :)

Answer 2

no doubts? :)`