IB Maths AI HL Scaling and Linearising Data Questions

1. Take the logarithm of both sides of the equation: ${\log }_{10}y={\log }_{10}(a{x}^n)$.

2. Apply the product rule for logarithms ($\log (ab)=\log a+\log b$): ${\log }_{10}y={\log }_{10}a+{\log }_{10}{x}^n$.

3. Apply the power rule for logarithms ($\log (x^n)=n\log x$): ${\log }_{10}y=n{\log }_{10}x+{\log }_{10}a$.

4. Map this to the linear equation $Y=mX+c$ to finish the proof. Here, $Y={\log }_{10}y$ and $X={\log }_{10}x$. The gradient is $m=n$ and the $y$-intercept is $c={\log }_{10}a$.

1. Linearise the exponential model by taking the natural logarithm of both sides: $\ln y=\ln a+x\ln b$.

2. Identify the $y$-intercept from the linearised form: $\ln a=5.1$.

3. Solve for $a$ by raising $e$ to both sides: $a=e^{5.1}$. This evaluates to $164.021\cdots \approx$ $164$.

4. Identify the gradient from the linearised form: $\ln b=-0.45$.

5. Solve for $b$ by raising $e$ to both sides: $b=e^{-0.45}$. This evaluates to $0.63762\cdots \approx$ $0.638$.

1. Substitute $Q=4$ directly into the regression equation: ${\log }_{10}P=2.5{\log }_{10}(4)-1.2$.

2. Calculate the value on the right-hand side using your GDC: $2.5(0.60205\dots )-1.2=1.50514\cdots -1.2$.

3. Simplify the result: ${\log }_{10}P=0.30514\dots$

4. Convert from logarithmic form back to exponential form to isolate $P$: $P={10}^{0.30514\dots }$.

5. Evaluate the final result to 3 significant figures: $P\approx 2.019\dots ⟹$ $2.02$.