The problem given is:

For which values of $a$ does the equation $9^x + 2 \cdot 3^{x+2} = a$ have real solutions? Solve the equation for these values of $a$.

Solution

Let's simplify and analyze the equation to determine the values of $a$.

1. Rewrite the terms in terms of $3^x$: $9^x = (3^2)^x = (3^x)^2$ This means we can substitute $y = 3^x$, so $9^x = y^2$.

   The equation then becomes: $y^2 + 2 \cdot y \cdot 3^2 = a$

2. Simplify further: Since $3^2 = 9$, we have: $y^2 + 2 \cdot y \cdot 9 = a$ $y^2 + 18y = a$

3. Determine values of $a$ for real solutions: For real values of $x$, $y = 3^x$ must also be real and positive (since $3^x$ is always positive for real $x$). We want to express $a$ in terms of $y$ and examine the resulting function: $a = y^2 + 18y$

4. Analyze the range of $a$: This expression $y^2 + 18y$ is a quadratic function of $y$ and opens upwards. The minimum value occurs at the vertex, where $y = -\frac{18}{2} = -9$.

   Substituting $y = -9$: $a = (-9)^2 + 18 \cdot (-9) = 81 - 162 = -81$

   So, the minimum value of $a$ is $-81$, and $a$ can take any value greater than or equal to $-81$.

Conclusion

The equation $9^x + 2 \cdot 3^{x+2} = a$ has real solutions for $a \geq -81$.

Would you like further clarification, or do you have any specific questions?

Related Questions

1. How can we find the exact values of $x$ for specific values of $a$?
2. What happens to the function $9^x + 2 \cdot 3^{x+2}$ as $x \to \infty$?
3. Could we use a graphical approach to confirm the minimum value of $a$?
4. How does the substitution $y = 3^x$ help in solving similar exponential equations?
5. What is the significance of the minimum value $a = -81$ in real-world applications?

Tip

When dealing with exponential equations, substituting a new variable for exponential terms can greatly simplify the process.