TMUA practice: Exponentials Logs, problem 3

Substitute $u = 2^{x}$. Then $4^{x} = (2^{x})^{2} = u^{2}$, so the equation becomes a quadratic in $u$:

$u^{2} - 5u + 4 \;=\; 0.$

Factorise: $\;(u - 1)(u - 4) = 0$, giving $u = 1$ or $u = 4$.

Return to $x$ via $u = 2^{x}$.

• $u = 1$: $\;2^{x} = 1$ gives $x = 0$.
• $u = 4$: $\;2^{x} = 4 = 2^{2}$ gives $x = 2$.

Both values are valid (the substitution $u = 2^{x}$ is well-defined for all real $x$, and $u > 0$ is automatic). The solution set is $\{0,\, 2\}$. This is option D.

Why the other options fail.

• A, $\{0\}$, has stopped after the first root $u = 1$.
• B, $\{2\}$, has stopped after the second root $u = 4$.
• C, $\{1,\, 4\}$, has reported the values of $u$ as if they were the values of $x$; the student forgot to undo the substitution.
• E, $\{-2,\, 2\}$, has wrongly assumed that the equation is symmetric under $x \mapsto -x$ (it is not: $4^{-x} \neq 4^{x}$ unless $x = 0$).

The lesson: an equation of the form $\;a^{2x} + b\cdot a^{x} + c = 0\;$ becomes a standard quadratic under the substitution $u = a^{x}$. After solving for $u$, remember that $a^{x} > 0$ always, so any negative or zero root for $u$ must be rejected, and remember to undo the substitution.