Question. Simplify $\left(\dfrac{p^{12}}{81q^{16}}\right)^{-3/4}$ and express with positive indices.

Solution.

Step 1 — Flip the fraction to handle the negative power:

$$\left(\dfrac{p^{12}}{81q^{16}}\right)^{-3/4} = \left(\dfrac{81q^{16}}{p^{12}}\right)^{3/4}$$

Step 2 — Distribute the $\tfrac{3}{4}$ to each factor:

$$= \frac{81^{3/4} \cdot (q^{16})^{3/4}}{(p^{12})^{3/4}}$$

Step 3 — Evaluate each piece.

• $81 = 3^4$, so $81^{3/4} = 3^{4 \cdot 3/4} = 3^3 = 27$.
• $(q^{16})^{3/4} = q^{16 \cdot 3/4} = q^{12}$.
• $(p^{12})^{3/4} = p^{12 \cdot 3/4} = p^9$.

Step 4 — Combine:

$$= \frac{27\,q^{12}}{p^9}$$

Answer: $\dfrac{27\,q^{12}}{p^9}$.

Lesson: negative outer power $\Rightarrow$ flip the fraction first; then distribute the rational power and simplify any perfect prime-powers like $81 = 3^4$.

Question. Simplify $\dfrac{(2xy)^{-2}}{12} \div \dfrac{x^3}{4y^2}$.

Solution.

Step 1 — Expand $(2xy)^{-2}$:

$$(2xy)^{-2} = 2^{-2}\,x^{-2}\,y^{-2} = \frac{1}{4x^2 y^2}$$

Step 2 — Substitute into the LHS:

$$\frac{1/(4x^2 y^2)}{12} = \frac{1}{48 x^2 y^2}$$

Step 3 — Change ÷ to × with reciprocal:

$$\frac{1}{48 x^2 y^2} \times \frac{4y^2}{x^3} = \frac{4 y^2}{48\,x^2 y^2\, x^3}$$

Step 4 — Simplify (numbers: $\tfrac{4}{48} = \tfrac{1}{12}$; variables: $y^2 / y^2 = 1$; $x^2 \cdot x^3 = x^5$):

$$= \frac{1}{12\,x^5}$$

Answer: $\dfrac{1}{12 x^5}$.

Lesson: convert negative-power brackets to fractions first; then turn ÷ into × via reciprocal; cancel numerically and via index laws separately.

Question. Solve $4^{m-3} \times 8^{3m-1} = 4$.

Solution.

Step 1 — Common base 2.

• $4 = 2^2$, so $4^{m-3} = 2^{2(m-3)} = 2^{2m-6}$.
• $8 = 2^3$, so $8^{3m-1} = 2^{3(3m-1)} = 2^{9m-3}$.
• RHS: $4 = 2^2$.

Step 2 — Combine LHS using $a^p \times a^q = a^{p+q}$:

$$2^{2m - 6} \cdot 2^{9m - 3} = 2^{(2m - 6) + (9m - 3)} = 2^{11m - 9}$$

Step 3 — Equate to RHS:

$$2^{11m - 9} = 2^2$$

Step 4 — Equate exponents:

$$11m - 9 = 2 \Rightarrow 11m = 11 \Rightarrow m = 1$$

Answer: $m = 1$.

Lesson: rewrite everything in the smallest prime base; collapse to one power per side; equate exponents.

Question. Solve $2^a + 2^a + 2^a + 2^a = 32$.

Solution.

Step 1 — Combine like terms. Four copies of $2^a$:

$$4 \cdot 2^a = 32$$

Step 2 — Divide both sides by 4:

$$2^a = 8$$

Step 3 — Same base 2. $8 = 2^3$:

$$2^a = 2^3 \Rightarrow a = 3$$

Answer: $a = 3$.

Lesson: $a^x + a^x + a^x + a^x = 4 \cdot a^x$ — combine like-terms first, then solve as a same-base equation.

Question. Given $\dfrac{625^p}{5^q} = 125^r$, express $q$ in terms of $p$ and $r$.

Solution.

Step 1 — Common base 5.

• $625 = 5^4 \Rightarrow 625^p = 5^{4p}$.
• $125 = 5^3 \Rightarrow 125^r = 5^{3r}$.

Step 2 — Substitute:

$$\frac{5^{4p}}{5^q} = 5^{3r}$$

Step 3 — Quotient law:

$$5^{4p - q} = 5^{3r}$$

Step 4 — Equate exponents:

$$4p - q = 3r$$

Step 5 — Rearrange for $q$:

$$q = 4p - 3r$$

Answer: $q = 4p - 3r$.

Lesson: rewrite all bases as the same prime, then read the relationship straight off the exponents.

Question. A bacteria population is modelled by $N = 500 \times 4^t$, where $t$ is in hours. After how many hours will the population reach $32{,}000$?

Solution.

Step 1 — Set up:

$$500 \times 4^t = 32{,}000$$

Step 2 — Isolate the power:

$$4^t = \frac{32{,}000}{500} = 64$$

Step 3 — Common base 2. $4 = 2^2$, $64 = 2^6$:

$$2^{2t} = 2^6 \Rightarrow 2t = 6 \Rightarrow t = 3$$

Answer: $t = 3$ hours.

Lesson: isolate $a^t$ first; then convert both sides to the same prime base and equate exponents.

Question. Which is larger, $2^{500}$ or $5^{250}$?

Solution — match exponents.

Step 1 — Factor out 250 from each exponent.

$$2^{500} = (2^2)^{250} = 4^{250}$$ $$5^{250} = 5^{250}$$

Step 2 — Compare with the same exponent. Both are raised to the power 250. Since $5 > 4$:

$$5^{250} > 4^{250} = 2^{500}$$

Answer: $5^{250} > 2^{500}$.

Lesson: when you can rewrite two powers with the same exponent, comparison reduces to comparing bases — much easier than computing the actual numbers.

Question. The graph of $y = ka^x$ passes through $(0, 3)$ and $(3, 24)$. Find $k$ and $a$.

Solution.

Step 1 — Use $(0, 3)$. When $x = 0$: $y = ka^0 = k \cdot 1 = k$. So $k = 3$.

Step 2 — Use $(3, 24)$ with $k = 3$:

$$24 = 3 \cdot a^3$$ $$a^3 = 8$$ $$a = 2$$

Answer: $k = 3$, $a = 2$. So $y = 3 \cdot 2^x$.

Check: at $x = 3$, $y = 3 \cdot 8 = 24$. ✓

Lesson: the $y$-intercept gives $k$ for free (since $a^0 = 1$); then one other point pins down $a$.