Reducing y = Ab^x to Linear Form

“Reduce to linear form” means taking logs so a curved relationship becomes a straight line $Y = mX + c$. Then the gradient and intercept give the constants. For $y = Ab^x$ the right move is $\lg$ of both sides, which is linear in $x$.

Two variables $x$ and $y$ are related by $y = Ab^x$, where $A$ and $b$ are constants. When $\lg y$ is plotted against $x$, a straight line of gradient $0.5$ and intercept $1$ (on the $\lg y$ axis) is obtained. Find the value of $A$ and the value of $b$. [6]

The working

Step 1, take $\lg$ of the model and use the log laws: $\lg y = \lg\left(A b^x\right) = \lg A + \lg\left(b^x\right) = \lg A + x\lg b \quad \text{(M1)}$

Step 2, match to $Y = mX + c$ with $Y = \lg y$ and $X = x$: $\underbrace{\lg y}_{Y} = \underbrace{(\lg b)}_{\text{gradient}} x + \underbrace{\lg A}_{\text{intercept}} \quad \text{(M1)}$

So the gradient is $\lg b$ and the intercept is $\lg A$.

Step 3, use the given gradient $0.5$: $\lg b = 0.5 \;\Rightarrow\; b = 10^{0.5} = \sqrt{10} \approx 3.16 \quad \text{(M1, A1)}$

Step 4, use the given intercept $1$: $\lg A = 1 \;\Rightarrow\; A = 10^{1} = 10 \quad \text{(M1, A1)}$

So $A = 10$ and $b = \sqrt{10} \approx 3.16$, giving $y = 10\left(\sqrt{10}\right)^x$.

Where the marks are won and lost

• Gradient $= \lg b$ and intercept $= \lg A$ is the whole idea. Swapping them (setting $\lg b = 1$) is the classic error, keep the coefficient of $x$ as the gradient.
• Undoing $\lg$ means raising $10$ to the power (not $e$). $\lg b = 0.5 \Rightarrow b = 10^{0.5}$.
• $A$ and $b$ are the original constants, so both need converting back from their logs. Leaving $\lg A = 1$ as the answer is incomplete.

Common mistakes

• Using $\ln$ (base $e$) to invert $\lg$ (base $10$).
• Reading the intercept as $b$ and the gradient as $A$.
• Forgetting to convert: reporting $\lg A$ and $\lg b$ instead of $A$ and $b$.

Full method: Reducing Relationships to Linear Form notes. See also Converting to Linear Form. Topic home: Logs & Exponentials pillar.