Mathematics 9709/22 — May/June 2014

Walkthrough

We need to solve $|x + 2| = |x - 13|$. The absolute value (modulus) of a number is its distance from zero, so $|a| = |b|$ means $a$ and $b$ are the same distance from zero, i.e. $a = b$ or $a = -b$.

A clean way to handle this is to square both sides, since squaring removes the absolute value: $|a|^2 = a^2$. So we get $(x + 2)^2 = (x - 13)^2$.

Expanding gives $x^2 + 4x + 4 = x^2 - 26x + 169$. The $x^2$ terms cancel, leaving the linear equation $4x + 4 = -26x + 169$. Adding $26x$ to both sides gives $30x + 4 = 169$, so $30x = 165$, and therefore $x = \frac{165}{30} = \frac{11}{2} = 5.5$.

We can verify: $|5.5 + 2| = |7.5| = 7.5$ and $|5.5 - 13| = |-7.5| = 7.5$. The two sides match, confirming the solution.

Key Takeaways

• Squaring both sides is a reliable way to eliminate absolute values in an equation.
• The equation $|a| = |b|$ is equivalent to $a^2 = b^2$ for real numbers $a$ and $b$.
• After squaring, the resulting equation is often simpler (here, linear).

Common Mistakes

• Forgetting to square the entire expression — writing $(x + 2)^2 = x - 13$ instead of $(x - 13)^2$.
• Sign errors when expanding $(x - 13)^2$: it is $x^2 - 26x + 169$, not $x^2 + 26x + 169$.
• Not simplifying $\frac{165}{30}$ to $\frac{11}{2}$.

Things to Be Careful About

• Squaring both sides is valid here because both sides are non-negative (they are absolute values).
• The alternative method is to write $x + 2 = -(x - 13)$, giving $x + 2 = -x + 13$, so $2x = 11$ and $x = \frac{11}{2}$. This works because the case $x + 2 = x - 13$ is impossible ($2 = -13$).

Walkthrough

The equation is $|3^y + 2| = |3^y - 13|$. This has exactly the same structure as $|x + 2| = |x - 13|$ from part (i), with $x$ replaced by $3^y$. Since we already found that $|x + 2| = |x - 13|$ has solution $x = \frac{11}{2}$, we can directly substitute: $3^y = \frac{11}{2}$.

Now we need to solve for $y$, which is in the exponent. The way to bring an exponent down is to take logarithms of both sides. Using base-10 logarithms (any consistent base works):

$\log(3^y) = \log\left(\frac{11}{2}\right)$

By the power law of logarithms, $\log(3^y) = y \log 3$. So:

$y \log 3 = \log\left(\frac{11}{2}\right)$

Dividing both sides by $\log 3$:

$y = \frac{\log(11/2)}{\log 3} \approx \frac{1.70475}{1.09861} \approx 1.5517$

Rounded to 3 significant figures, $y = 1.55$.

Key Takeaways

• The word "hence" signals that the previous result should be reused directly.
• To solve for an unknown in the exponent, take logarithms of both sides and use the power law.
• The base of the logarithm does not matter as long as it is consistent — the ratio $\frac{\log a}{\log b}$ is the same in any base.

Common Mistakes

• Not recognising the substitution $x = 3^y$ from part (i).
• Misapplying the power law: $\log(3^y) = y \log 3$, not $(\log 3)^y$.
• Rounding too early — keep extra decimal places until the final step, then round to 3 s.f.

Things to Be Careful About

• The answer must be given to 3 significant figures: $1.55$, not $1.5$ or $1.5517$.
• Take logarithms of both sides consistently.
• $3^y$ is always positive, so $3^y + 2 > 0$ always; this does not affect the solution since we use the result from part (i) directly.