(1/7)^2x+6*(1/7)^x-7<0

Let's analyze and solve the inequality: \[ \left(\frac{1}{7}\right)^{2x} + 6 \cdot \left(\frac{1}{7}\right)^x - 7 < 0 \] --- ### Step 1: Make a substitution Let: \[ t = \left(\frac{1}{7}\right)^x \] Since \( \frac{1}{7} > 0 \), then for any real \( x \): - \( t > 0 \) - And because \( \left(\frac{1}{7}\right)^x = 7^{-x} \), it is decreasing for increasing \( x \). In terms of \( t \), the inequality becomes: \[ t^2 + 6t - 7 < 0 \] --- ### Step 2: Solve the quadratic inequality Factor or use the quadratic formula for: \[ t^2 + 6t - 7 < 0 \] Discriminant: \[ \Delta = 6^2 - 4 \cdot 1 \cdot (-7) = 36 + 28 = 64 \] Roots: \[ t = \frac{-6 \pm \sqrt{64}}{2} = \frac{-6 \pm 8}{2} \] Calculating roots: - \( t_1 = \frac{-6 - 8}{2} = \frac{-14}{2} = -7 \) - \( t_2 = \frac{-6 + 8}{2} = \frac{2}{2} = 1 \) Since the quadratic coefficient (a=1) is positive, the parabola opens upward. The inequality \( t^2 + 6t - 7 < 0 \) holds between the roots: \[ -7 < t < 1 \] --- ### Step 3: Recall the domain of \( t \) We have: \[ t = \left(\frac{1}{7}\right)^x > 0 \] Also, since \( t \) is always positive: \[ t > 0 \] From the interval above: \[ -7 < t < 1 \] Intersecting with \( t > 0 \), the solution for \( t \) is: \[ 0 < t < 1 \] ### Step 4: Convert back to \( x \) Recall: \[ t = \left(\frac{1}{7}\right)^x \] The inequality: \[ 0 < \left(\frac{1}{7}\right)^x < 1 \] Since \( \frac{1}{7} \) is between 0 and 1: - \( \left(\frac{1}{7}\right)^x < 1 \) for all \( x \), - \( \left(\frac{1}{7}\right)^x > 0 \) for all real \( x \). Now, to find the values of \( x \) such that: \[ \left(\frac{1}{7}\right)^x < 1 \] which is always true, *but* considering the bounds: \[ \left(\frac{1}{7}\right)^x < 1 \] implies: \[ \left(\frac{1}{7}\right)^x = 1 \Rightarrow x = 0 \] and for: \[ \left(\frac{1}{7}\right)^x < 1 \] since \( \frac{1}{7} < 1 \), the function decreases as \( x \) increases. From the inequality \( 0 < t < 1 \), we get: \[ \left(\frac{1}{7}\right)^x < 1 \] which holds for: \[ x > 0 \] and \[ \left(\frac{1}{7}\right)^x > 0 \] for all \( x \). --- ### **Final Answer:** \[ \boxed{ \text{The solution set is } \quad x > 0 } \] --- If you'd like, I can also provide a more detailed explanation or check for boundary points!