(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!
