LAWS OF ALGEBRA

QUESTION 1 [6 marks] — Criterion A Medium

Rewrite each with a positive index, then evaluate:

a. $2^{-3}$

[2]

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$$2^{-3}=\frac{1}{2^3}=\frac{1}{8}$$

b. $5^{-2}$

[2]

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$$5^{-2}=\frac{1}{5^2}=\frac{1}{25}$$

c. $4 \times 3^{-1}$

[2]

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$$4\times\frac{1}{3}=\frac{4}{3}$$

QUESTION 2 [4 marks] — Criterion B Medium

Investigate the connection between the negative index law and the division index law, using $\dfrac{x^2}{x^5}$.

a. Simplify $\dfrac{x^2}{x^5}$ using the division law $\dfrac{x^m}{x^n}=x^{m-n}$.

[1]

Show Solution

$$x^{2-5}=x^{-3}$$

b. Now simplify $\dfrac{x^2}{x^5}$ by cancelling common factors directly (writing out the $x$'s).

[2]

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$\dfrac{x\times x}{x\times x\times x\times x\times x} = \dfrac{1}{x\times x\times x}=\dfrac{1}{x^3}$

c. Since both methods calculate the same thing, what must $x^{-3}$ equal?

[1]

Show Solution

$$x^{-3}=\frac{1}{x^3}$$

QUESTION 3 [4 marks] — Criterion C Medium

A student simplifies $2^{-3}$ as $-8$ (treating the negative sign as making the whole answer negative).

a. Explain the error, and give the correct value.

[2]

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A negative EXPONENT does not make the result negative — it means 'reciprocal' (1 over the positive-power version). Correct: $2^{-3}=\frac{1}{2^3}=\frac{1}{8}$, not $-8$.

b. Explain how $2^{-3}$ (positive result) is different from $-(2^3)$ (which genuinely does give a negative result).

[2]

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$2^{-3}=\frac{1}{8}$ (positive, since the negative applies to the EXPONENT, meaning 'take the reciprocal'). $-(2^3)=-8$ (negative, since the minus sign applies to the whole result AFTER evaluating $2^3=8$). These are two completely different operations that happen to use similar-looking negative signs.

QUESTION 4 [3 marks] — Criterion D Medium

In physics, the intensity of light follows an inverse-square law: intensity is proportional to $d^{-2}$, where $d$ is distance from the source.

a. Rewrite $d^{-2}$ using a positive index (as a fraction).

[1]

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$$d^{-2}=\frac{1}{d^2}$$

b. Explain, using this rewritten form, why doubling the distance from a light source (i.e. $d\to2d$) reduces the intensity to $\frac{1}{4}$ of its original value, not $\frac{1}{2}$.

[2]

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Since intensity $\propto\frac{1}{d^2}$, doubling $d$ gives $\frac{1}{(2d)^2}=\frac{1}{4d^2}=\frac{1}{4}\times\frac{1}{d^2}$ — the intensity is divided by $2^2=4$, not just 2, because the distance is SQUARED in the denominator.

QUESTION 5 [4 marks] — Criterion A Medium

Rewrite each with a positive index, then evaluate as a fraction:

a. $6^{-2}$

[2]

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$$\frac{1}{6^2}=\frac{1}{36}$$

b. $5 \times 2^{-3}$

[2]

Show Solution

$$5\times\frac{1}{8}=\frac{5}{8}$$

QUESTION 6 [3 marks] — Criterion A Medium

Simplify $\dfrac{1}{5^{-2}}$ (a NEGATIVE index in the DENOMINATOR).

a. Rewrite $5^{-2}$ with a positive index first, then simplify the overall fraction (dividing by a fraction).

[3]

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$$\frac{1}{5^{-2}} = \frac{1}{\frac{1}{25}} = 1\div\frac{1}{25} = 1\times25=25=5^2$$

QUESTION 7 [5 marks] — Criterion B Hard

Investigate the general rule for $\dfrac{1}{x^{-n}}$, based on the specific case $\dfrac{1}{5^{-2}}=5^2$ found above.

a. Test the pattern with $\dfrac{1}{3^{-4}}$ — does it equal $3^4$? Verify numerically.

[2]

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$3^{-4}=\frac{1}{81}$. $\frac{1}{3^{-4}}=\frac{1}{\frac{1}{81}}=81=3^4$ ?, confirming the pattern.

b. State the general rule for $\dfrac{1}{x^{-n}}$ in terms of $x^n$, and explain WHY 'a negative index in the denominator flips to become a positive index' using the definition of a negative index as a reciprocal.

[3]

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$$\frac{1}{x^{-n}} = x^n$$ Since $x^{-n}$ ALREADY means $\frac{1}{x^n}$ (its definition as a reciprocal), taking the reciprocal of THAT (i.e. $\frac{1}{x^{-n}}$) means finding the reciprocal of a reciprocal, which always returns to the original value — hence $\frac{1}{x^{-n}}=\frac{1}{\frac{1}{x^n}}=x^n$.

QUESTION 8 [4 marks] — Criterion C Medium

A student simplifies $3^{-2}$ as $-9$ (treating the negative index as making the base negative before squaring).

a. Explain the error, and give the correct value, distinguishing between $3^{-2}$, $(-3)^2$, and $-(3^2)$.

[4]

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$3^{-2}$ means 'take the reciprocal of $3^2$', giving $\frac{1}{9}$ (positive, since it's a fraction, not related to sign at all). This is completely different from $(-3)^2=9$ (squaring a negative base gives positive) and $-(3^2)=-9$ (negating the result of squaring). The student incorrectly conflated 'negative index' with 'negative base' — these are unrelated concepts that happen to both involve the word 'negative'.

QUESTION 9 [6 marks] — Criterion D Hard

The intensity of sound decreases with distance according to $I \propto d^{-2}$, similar to light intensity. A sound source has intensity 80 units at 1m distance.

a. Using $I=\dfrac{80}{d^2}$, find the intensity at $d=4$m.

[2]

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$$I=\frac{80}{4^2}=\frac{80}{16}=5 \text{ units}$$

b. A safety guideline requires intensity below 2 units. Find the MINIMUM distance required to meet this guideline, and explain (using the inverse-square relationship) why doubling the required 'safe intensity' threshold (e.g. from 2 to 4 units) would NOT simply double the minimum safe distance.

[4]

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$\frac{80}{d^2}\le2 \Rightarrow d^2\ge40 \Rightarrow d\ge\sqrt{40}\approx6.32$m. Because intensity depends on $d^{-2}$ (an INVERSE SQUARE relationship, not a simple inverse), changing the threshold doesn't scale distance linearly — doubling the allowed intensity threshold to 4 would give $d^2\ge20 \Rightarrow d\ge\sqrt{20}\approx4.47$m, which is NOT half of 6.32m; because of the squared relationship, distance changes by a factor of $\sqrt{2}$ (not 2) when intensity threshold changes by a factor of 2.

The negative index law