The equation \[8^x+4=4^x+2^{x+2}\] has

  1. no real solutions;

  2. one real solution;

  3. two real solutions;

  4. three real solutions.

[A real solution is one that lies on the number line.]

First set \(y=2^x\). Then the equation can be rewritten as \[\begin{align*} &y^3+4=y^2+4y\\ \iff &y^3-y^2-4y+4=0. \end{align*}\] The left hand side can be factorised (by noticing that if we substitute \(y=1\) then we have a solution). \[\begin{align*} &(y-1)(y^2-4)=0\\ \iff &(y-1)(y-2)(y+2)=0 \end{align*}\]

So the possible solutions are \(y=1\), \(y=2\), \(y=-2\). But \(y=2^x>0\) for real \(x\), so valid solutions occur when \(y=1\) or \(y=2\).

This gives the solutions \(x=0\) and \(x=1\) respectively, but since we’re only asked how many solutions there are, we don’t actually need to find them (provided we can be sure they exist!)

So the equation has two real solutions. The answer is (c).

The curve whose roots are those required.
The curve \(y=8^x+4-4^x-2^{x+2}\)