Revise Exponents and logarithms for IB Mathematics: Applications and Interpretation (1.4)

(a) Combine using the product rule on top, then the quotient rule:\ $\dfrac{x^7 \times x^2}{x^3} = \dfrac{x^{7+2}}{x^3} = x^{9-3} = x^{6}$. (M1 for adding/subtracting indices) (A1)\ (b) Apply the power rule, then the negative index:\ $(2^3)^{-2} = 2^{3 \times (-2)} = 2^{-6} = \dfrac{1}{2^6} = \dfrac{1}{64}$. (A1)\ (c) A fractional index is a root then a power:\ $27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$. (A1)

1. Set up the equation with the target value:\ $1600 = 100 \times 2^{t}$.\ 2. Isolate the exponential by dividing by 100:\ $16 = 2^{t}$. (M1 for isolating the power)\ 3. Rewrite using the definition of a logarithm, $\log_a x = y \iff a^y = x$:\ $t = \log_2 16$. (M1 for converting to log form)\ Since $2^4 = 16$, $t = 4$.\ It takes 4 hours for the colony to reach 1600 bacteria. (A1)

1. Set up the equation:\ $10000 = 25000 \times (0.85)^{t}$.\ 2. Isolate the exponential by dividing by 25000:\ $0.4 = (0.85)^{t}$. (M1 for isolating the power)\ 3. Take the natural log of both sides:\ $\ln 0.4 = \ln\!\left((0.85)^{t}\right)$.\ 4. Apply the power rule to bring $t$ down:\ $\ln 0.4 = t \ln 0.85$. (M1 for taking logs and using the power rule)\ 5. Rearrange and evaluate:\ $t = \dfrac{\ln 0.4}{\ln 0.85} = 5.6369\ldots$ (M1 for rearranging for $t$)\ So $t \approx 5.64$ years (3 s.f.). (A1)\ On Paper 2 you could instead plot $y = 25000(0.85)^x$ and $y = 10000$ on the GDC and read the intersection, $x = 5.64$ — equally valid.

Model answer — full working.\ \ Isolate the power. Divide both sides by 500:\ $1.04^{t} = \dfrac{800}{500} = 1.6.$\ \ Take logs of both sides. Using the natural log:\ $\ln\!\left(1.04^{t}\right) = \ln 1.6.$\ \ Bring the exponent down with the power rule and rearrange for $t$:\ $t \ln 1.04 = \ln 1.6 \quad\Rightarrow\quad t = \dfrac{\ln 1.6}{\ln 1.04} = 11.987\ldots$\ \ Round. $t \approx 12.0$ (3 s.f.).\ \ Alternative (GDC): plot $y = 500 \times 1.04^{x}$ and $y = 800$ and find the intersection, or use the equation solver, giving $x = 12.0$ — fully acceptable on Paper 2.\ \ ---\ How our marking engine awards the 4 marks:\ \ - M1 — isolate the power. Awarded for reaching $1.04^{t} = 1.6$ (dividing both sides by 500). This is the approach mark, so it survives an arithmetic slip further down.\ - M1 — take logs (or GDC solve). Awarded for applying $\ln$ (or $\log$) to both sides, $\ln(1.04^{t}) = \ln 1.6$ — or for setting up a valid GDC solution (graph intersection or solver). The engine accepts either route.\ - M1 — rearrange for $t$. Awarded for using the power rule and isolating the variable: $t = \dfrac{\ln 1.6}{\ln 1.04}$. This depends on a valid log step above, and FT applies — a candidate who isolated a slightly different power but rearranged correctly still earns it on their own figures.\ - A1 — final value. Awarded for $t \approx 12.0$, correctly rounded to 3 significant figures. This accuracy mark depends on the method marks that precede it.\ \ 'Accept equivalent forms and correct rounding.' The engine accepts $\dfrac{\ln 1.6}{\ln 1.04}$, $\dfrac{\log 1.6}{\log 1.04}$, $11.99$ carried to more figures, or the GDC value 12.0; it accepts the answer whether reached by logs or graphically. Once the correct value appears, ISW (ignore subsequent working) means a later restatement does not lose marks.\ \ Bottom line: three of the four marks are method marks that survive an arithmetic slip, and the single accuracy mark is shielded by follow-through. A student who writes only '$t = 12$' with no working risks losing the three method marks; a student who shows the isolation, the log step and the rearrangement keeps the method regardless of a slip in the final figure.