When does $9^x - 3^{x+1} = k$ have one or more real solutions?

When does $9^x - 3^{x+1} = k$ have one or more real solutions?

Question

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

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

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1. $k > 0$,

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1. $k \leq -1$,

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1. $k \geq 5/8$.

Suggestion

Maybe we can change variable so that we are working with an equation of a more familiar form? Would the substitution $y = 3^x$ help?

Solution

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).

Interactive check

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/TDQ9jTyn/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

The green curve is $y=9^x - 3^{x+1} -k$.