Question 1209741
Here's how to solve the equation 1/9^(1/x) + 1/3^(1/x) = 30:

1. **Rewrite with a common base:** Notice that 9 = 3². We can rewrite the first term as:

   1/9^(1/x) = 1/(3²)^(1/x) = 1/3^(2/x) = (1/3^(1/x))²

2. **Substitute:** Let y = 1/3^(1/x).  The equation becomes:

   y² + y = 30

3. **Rearrange:**

   y² + y - 30 = 0

4. **Factor:**

   (y + 6)(y - 5) = 0

5. **Solve for y:**

   y = -6 or y = 5

6. **Consider the valid solution:** Since y = 1/3^(1/x), y must be positive. Therefore, y = -6 is not a valid solution.  We are left with:

   y = 5

7. **Substitute back:**

   1/3^(1/x) = 5

8. **Rewrite:**

   3^(-1/x) = 5

9. **Take the logarithm of both sides (any base will work, but natural log is common):**

   ln(3^(-1/x)) = ln(5)

10. **Use logarithm power rule:**

   (-1/x) * ln(3) = ln(5)

11. **Solve for x:**

   -1/x = ln(5) / ln(3)
   1/x = -ln(5) / ln(3)
   x = -ln(3) / ln(5)

12. **Calculate:**

   x ≈ -1.465 / 1.609
   x ≈ -0.911

Therefore, x ≈ -0.911.