9^x+2×3^x-117/3^x-27≤1

To solve this inequality, we can first simplify the expression by factoring out a common factor of 3^x:

9^x + 2*3^x - 117 / 3^x - 27

= 3^(2x) + 2*3^x - 117 / 3^x - 3^3

= 3^x(3^x + 2) - 117 / 3^x - 27

Now, we can rewrite the inequality as:

(3^x(3^x + 2) - 117) / (3^x - 27) ≤ 1

Next, we can multiply both sides of the inequality by (3^x - 27) to get rid of the denominator:

3^x(3^x + 2) - 117 ≤ 3^x - 27

Expand the left side:

3^(2x) + 2*3^x - 117 ≤ 3^x - 27

Now, let y = 3^x, then we have:

y^2 + 2y - 117 ≤ y - 27

Rearrange the inequality:

y^2 + y - 90 ≤ 0

Now, we can factor the quadratic equation:

(y + 10)(y - 9) ≤ 0

This gives us the solutions y ≤ -10 or y ≥ 9.

Since y = 3^x, we can substitute back to get the solutions for x:

3^x ≤ -10 or 3^x ≥ 9

The exponential function is always positive, so there are no real solutions for 3^x ≤ -10.

For 3^x ≥ 9, we have:

x ≥ log3(9) = 2

Therefore, the solution to the inequality is x ≥ 2.