Review question

When does $9^x - 3^{x+1} = k$ have one or more real solutions? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7384

Solution

The equation $9^x - 3^{x+1} = k$ has one or more real solutions precisely when

1. $k \geq -9/4$,

2. $k > 0$,

3. $k \leq -1$,

4. $k \geq 5/8$.

If we write $y = 3^x$ and use the fact that $9^x = (3^2)^x = (3^x)^2$, the equation can be written as $y^2 -3y - k = 0.$

This has solutions $y = \frac{3 \pm \sqrt{9 + 4k}}{2}$ which are real when $k \geq -9/4$. Since $y = 3^x$, we need at least one of these solutions to be positive for $x$ to be real, but the larger root is clearly always positive, so this is not a problem. Hence the answer is (a).