how do i solve this? 9^2x=27^x-1?

Answer 1

9^(2x) = 27^(x - 1)
81^x = 27^(x - 1)
(27^x)(3^x) = 27^(x - 1)
3^x = 27^(x - 1 - x)
3^x = 3^(-3)

x = -3

Answer 2

What you want to do is establish 3 as your base, since both 9 and 27 are powers of 3. So, 9^2x = 3^[2(2x)], and 27^(x-1) = 3^[3(x-1)]. So, 3^[2(2x)] = 3^[3(x-1)]. At this point, just solve
2(2x) = 3(x-1). 4x = 3x-3, and x = -3.

Answer 3

Take Ln() or Log() in both sides of the equation:

9^2x=27^x-1

2x Ln(9) = Ln(27^x)/Ln(1)

2x = x Ln(27)/(Ln(1) * Ln(9))

x = 1/(2- Ln(27)/Ln(10))

x = 1/(2 - Ln(7))

Use the properties of logaritms.

Note: The difference between this procedure, and the one that it's above it that he took 27^x-1=27^(x-1). And I took 27^x-1 = (27^x) - 1

Answer 4

9^(2x) = 27^(x -1)
9 =3^2 and 27 = 3^3
[3^2]^2x = [3^3]^(x - 1)
3^(2*2x) = 3^(3*(x - 1))
3^4x = 3^(3x - 3)
both sides are of the same base, so
4x = 3x - 3
x = -3