9^x-1 -10*3^x-2 +1=0

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To solve for x in the given equation 9^x-1 -10*3^x-2 +1=0, let's break it down into two separate equations:

Let's rewrite the equation as follows:

(3^2)^(x-1) - 10(3^x)/(3^2) +1 = 0
3^(2x-2) - 103^(x-2) + 1 = 0

Let u = 3^(x-2)

Now, the equation becomes:

u^2 - 10u + 1 = 0

To solve this quadratic equation, we can use the quadratic formula:

u = [-(-10) ± sqrt((-10)^2 - 411)] / (2*1)
u = [10 ± sqrt(100 - 4)] / 2
u = [10 ± sqrt(96)] / 2
u = [10 ± 4√6] / 2

So, u = 5 ± 2√6

Now, recall that u = 3^(x-2), we can solve for x as follows:

Case 1: u = 5 + 2√6
3^(x-2) = 5 + 2√6
x-2 = log_3(5 + 2√6)
x = log_3(5 + 2√6) + 2

Case 2: u = 5 - 2√6
3^(x-2) = 5 - 2√6
x-2 = log_3(5 - 2√6)
x = log_3(5 - 2√6) + 2

Therefore, the solutions for x are:
x = log_3(5 + 2√6) + 2 and x = log_3(5 - 2√6) + 2.