1. **Stating the problem:** We have the compound interest formula: $$F = P(1 + i)^t$$ We need to make each of the following the subject of the formula: (a) $P$ (b) $i$ (c) $t$ 2. **Rearranging for (a) $P$:** Start with: $$F = P(1 + i)^t$$ Divide both sides by $(1 + i)^t$ to isolate $P$: $$P = \frac{F}{(1 + i)^t}$$ Intermediate step showing cancellation: $$P = \frac{F}{\cancel{(1 + i)^t}} \times \cancel{\frac{1}{(1 + i)^t}}$$ 3. **Rearranging for (b) $i$:** Start with: $$F = P(1 + i)^t$$ Divide both sides by $P$: $$\frac{F}{P} = (1 + i)^t$$ Intermediate step: $$\frac{\cancel{F}}{\cancel{P}} = (1 + i)^t$$ Take the $t$-th root (or raise both sides to the power $\frac{1}{t}$): $$\left(\frac{F}{P}\right)^{\frac{1}{t}} = 1 + i$$ Subtract 1 from both sides: $$i = \left(\frac{F}{P}\right)^{\frac{1}{t}} - 1$$ 4. **Rearranging for (c) $t$:** Start with: $$F = P(1 + i)^t$$ Divide both sides by $P$: $$\frac{F}{P} = (1 + i)^t$$ Take the natural logarithm of both sides: $$\ln\left(\frac{F}{P}\right) = \ln\left((1 + i)^t\right)$$ Use logarithm power rule: $$\ln\left(\frac{F}{P}\right) = t \ln(1 + i)$$ Divide both sides by $\ln(1 + i)$: $$t = \frac{\ln\left(\frac{F}{P}\right)}{\ln(1 + i)}$$ Intermediate step showing cancellation: $$t = \frac{\ln\left(\frac{F}{P}\right)}{\cancel{\ln(1 + i)}} \times \cancel{\frac{1}{\ln(1 + i)}}$$ **Final answers:** (a) $$P = \frac{F}{(1 + i)^t}$$ (b) $$i = \left(\frac{F}{P}\right)^{\frac{1}{t}} - 1$$ (c) $$t = \frac{\ln\left(\frac{F}{P}\right)}{\ln(1 + i)}$$