# Reduction to Linear Form

## Reduction to Linear Form Revision

**Reduction to Linear Form**

Some **exponential equations** can be reduced to a form that looks like y=mx+c. Specifically, after applying the **laws of logarithms** we can treat y=ax^{n} and y=ab^{x} as if they were **linear equations**.

## \mathbf{y=ax^{n}}

y=ax^{n}Take** logarithms**:

\begin{aligned}\log(y)&=\log(ax^{n})\\[1.2em]&=\log(a)+\log(x^{n})\\[1.2em]&=\log(a)+n\log(x)\end{aligned}

Overall we have:

y=ax^{n}\Rightarrow \log(y)=\log(a)+n\log(x)

Now if we plot \log(y) against \log(x), we have a **straight line graph**.

The graph below shows y against x in red and \log(y) against \log(x) in blue. As expected, the **blue line is straight**.

## \mathbf{y=ab^{x}}

y=ab^{x}Take **logarithms**:

\begin{aligned}\log(y)&=\log(ab^{x})\\[1.2em]&=\log(a)+\log(b^{x})\\[1.2em]&=\log(a)+x\log(b)\end{aligned}

Overall we have:

y=ab^{x}\Rightarrow \log(y)=\log(a)+x\log(b)

Now if we plot \log(y) against x, we have a **straight line graph**.

The graph below shows y against x in red and \log(y) against x in blue. As expected, the **blue line is straight**.

**Example 1: Converting to Linear Form**

Convert y=99\times0.5^{x} to **linear form**.

**[2 marks]**

y=99\times0.5^{x}

\begin{aligned}\log(y)&=\log(99\times0.5^{x})\\[1.2em]&=\log(99)+log(0.5^{x})\\[1.2em]&=\log(99)+x\log(0.5)\end{aligned}

**Example 2: Using Linear Form**

The data below is taken from a plot of the form y=ax^{n}. By plotting \log_{10}(y) against \log_{10}(x), find a and n.

**[5 marks]**

**Step 1: Calculate** \log_{10}(x) and \log_{10}y and **put them in a table**.

**Step 2: **Plot \log(y) against \log(x) on a **scatter graph**.

**Step 3: **Use the scatter graph to **find the gradient and y-intercept**.

Gradient is approximately \dfrac{3.3-0.6}{0.9-0}=\dfrac{2.7}{0.9}=3

y-intercept is approximately 0.6

**Step 4: **Use our **linear form** to **interpret the gradient and y-intercept**.

Gradient is n so n=3

y-intercept is \log(a) so \log(a)=0.6 so a=10^{0.6}=4 to two significant figures.

**Step 5: **Put together to **determine the form of the plot**:

## Reduction to Linear Form Example Questions

**Question 1: **The number of branches of a high-street store decreases over time. This trend, which is of the form y=ab^{x} is monitored over a number of years. Use the data from the monitoring to find a and b.

**[5 marks]**

If the question does not specify a base for logarithms it is up to us to choose a sensible base. The choice of base should not affect the answer at the end. In the working below, we have used a base of 2.

Note the linear form:

y=ab^{x}\rightarrow log(y)=x\log(b)+\log(a)So the straight line is obtained by plotting \log(y) against x.

This means that we need to add a \log(y) row to the table.

Plot \log(y) against x and add a line of best fit.

Find the gradient and the y-intercept.

Gradient is \dfrac{6-9.5}{6-0}=\dfrac{3.5}{6}=-\dfrac{7}{12}

y-intercept is 9.5

Interpret the gradient and y-intercept.

Gradient is \log(b)

\log(b)=-\dfrac{7}{12}

b=0.667

y-intercept is \log(a)

\log(a)=9.5

a=729

Put it all together:

Our estimate for the line is y=729\times0.667^{x}

**Question 2: **The value of an investment in MathCoin is believed to reliably climb following an ax^{n} trajectory. Claire buys one MathCoin and monitors the price of her investment every month over one year. Find a and n.

**[5 marks]**

If the question does not specify a base for logarithms it is up to us to choose a sensible base. The choice of base should not affect the answer at the end. In the working below, we have used a base of 10.

Note the linear form:

y=ax^{n}\rightarrow log(y)=n\log(x)+\log(a)So the straight line is obtained by plotting \log(y) against \log(x).

This means that we need to add a \log(x) row and a \log(y) row to the table.

Plot \log(y) against \log(x) and add a line of best fit.

Find the gradient and the y-intercept.

Gradient is \dfrac{3.5-2}{1-0}=\dfrac{1.5}{1}=1.5

y-intercept is 2

Interpret the gradient and y-intercept.

Gradient is n

n=1.5

y-intercept is \log(a)

log(a)=2

a=100

Put it all together:

Our estimate for the line is y=100\times x^{1.5}

**Question 3: **A forest has been declining in population since the nineteenth century. To assess this decline, a census of the tree population of the forest was conducted every ten years in the twentieth century. It is believed this decline follows a y=ab^{x} curve. Use the table of census data to find a and b.

**[5 marks]**

Again we can choose our own base for logarithms, so we will take the logarithm with base 3.

y=ab^{x}

\begin{aligned}\log(y)&=\log(ab^{x})\\[1.2em]&=\log(a)+\log(b^{x})\\[1.2em]&=\log(a)+x\log(b)\end{aligned}

So we plot \log(y) against x, and the gradient will be \log(b) and the y intercept will be \log(a).

Plotting \log(y) against x gives this graph.

The graph goes through the points (1,12.357) and (10,10).

\begin{aligned}\text{gradient}&=\dfrac{10-12.357}{10-1}\\[1.2em]&=\dfrac{-2.357}{9}\\[1.2em]&=-0.262\end{aligned}

\log(b)=-0.262

b=3^{-0.262}

b=0.750

The graph passes through (1,12.357) and has a gradient of -0.262, so it also passes through (0,12.357+0.262)=(0,12.619)

So y intercept is 12.619

\log(a)=12.619

a=3^{12.619}

a=1048576

Hence, y=1048576\times0.750^{x}

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