Mathematics 9709/32 — May/June 2018

Walkthrough

The key move is to introduce a temporary variable $y = 2^x$. This simplifies the appearance of the equation: $3|y - 1| = y$. The modulus means we must consider both signs inside the absolute value. When $y \ge 1$, the quantity $y - 1$ is already non-negative, so $|y - 1| = y - 1$. Solving $3(y - 1) = y$ gives $y = \frac{3}{2}$, which satisfies $y \ge 1$. When $y < 1$, the quantity is negative, so $|y - 1| = 1 - y$. Solving $3(1 - y) = y$ gives $y = \frac{3}{4}$, which satisfies $y < 1$. Both values of $y$ are positive, so both can be written as powers of $2$. To recover $x$ from $2^x = a$, take logs: $x = \frac{\ln a}{\ln 2}$. This gives $x = 0.5849625...$ and $x = -0.4150375...$, which round to $0.585$ and $-0.415$ at 3 significant figures.

Key Takeaways

This question combines absolute-value equations with exponential equations. When solving $|f(x)| = g(x)$, split into the cases $f(x) \ge 0$ and $f(x) < 0$, solve each separately, and check each candidate satisfies its case. It also uses the general fact that an equation of the form $a^x = b$ for $a, b > 0$ may be solved by logarithms.

Common Mistakes

A common mistake is to drop the modulus and solve only $3(2^x - 1) = 2^x$, which would miss the second solution. Another is to forget to check that an obtained value of $2^x$ lies in the assumed case. Also, students may fail to use logarithms correctly, or may state answers to insufficient accuracy; the question explicitly requires 3 significant figures.

Things to Be Careful About

The variable $y = 2^x$ is always positive, so any negative or zero value of $y$ would be invalid. Here both $\frac{3}{2}$ and $\frac{3}{4}$ are valid. When rounding, $0.5849625...$ rounds to $0.585$, and $-0.4150375...$ rounds to $-0.415$; note that the negative answer keeps three significant figures. Do not reject one of the two answers, because both satisfy the original equation.