Solve Log Base 1/7 Inequality: Finding Domain of x²+3x < 2log(3x+1)

To solve the inequality $\log_{\frac{1}{7}}(x^2+3x) < 2\log_{\frac{1}{7}}(3x+1)$, let's proceed step by step:

• Step 1: Simplify the right side using the power rule of logarithms:
  $2\log_{\frac{1}{7}}(3x+1) = \log_{\frac{1}{7}}((3x+1)^2)$.
• Step 2: The inequality becomes:
  $\log_{\frac{1}{7}}(x^2+3x) < \log_{\frac{1}{7}}((3x+1)^2)$.
• Step 3: Since the base of the logarithm is $\frac{1}{7}$, which is less than 1, the inequality changes direction:
  $x^2 + 3x > (3x + 1)^2$.
• Step 4: Expand and simplify:
  Expanding the right side: $x^2 + 3x > 9x^2 + 6x + 1$.
  
• Step 5: Rearrange the inequality:
  $0 > 8x^2 + 3x + 1$.
• Step 6: Attempt to solve $8x^2 + 3x + 1 < 0$:
  The discriminant of this quadratic, $(b^2 - 4ac)$, is $3^2 - 4 \times 8 \times 1 = 9 - 32 = -23$, which is less than 0. Therefore, the quadratic has no real roots.
• Step 7: Conclusion:
  The inequality $8x^2 + 3x + 1 < 0$ has no solution in terms of real $x$. The domain of $x$ satisfying this is an empty set.

Therefore, the solution is No solution.