Solve the equation 9^3x=81^x+1?

SHOULD be written as :-
9^(3x) = 81^(x + 1)
9^(3x) = 9^(2x + 2)
3x = 2x + 2
x = 2

9^(3x) = 81^(x+1)

9^(3x) = (9^2)^(x+1)

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

3x = 2x + 2

x = 2

3^2(3x)= 3^4(x+1)
3^6x= 3^(4x+4)
6x=4x + 4
2x=4
x=2

9^3x=81^x+1
(9)^3x=[(9)2]^x+1 (can write 81 as 9 square)
(9)^3x=(9)^2x+2 (multiply exponents)
since bases are same, exponents can be equated
3x=2x+2
3x-2x=2
x=2

9^3x = 9^2x+2
bases same so powers equal

3x=2x+2
3x-2x=2
x=2

9^3x=81^(x+1) can be re-written as:

9^3x=(9^2)^(x+1) or 9^3x=9^2(x+1) Since the two sides now have the same base, we can set the exponents equal.

3x = 2(x + 1) = 2x + 2 Subtracting 2x from both sides gives

x = 2

x=2

9^3x=9^2(x+1)...............81=9^2

so
3x=2x+2
x=2

9^3x = 81^x+1
27x = 81x + 1
27x - 81x = 81x - 81x + 1
-56x = 1
-1x = 1/56
x = -1/56
x = 0.017857142

9^(3x) = 81^(x + 1)
(3^2)^(3x) = (3^4)^(x + 1)
3^[2(3x)] = 3^[4(x + 1)]
2(3x) = 4(x + 1)
6x = 4x + 1
6x - 4x = 1
2x = 1
x = 1/2 (0.5)