# Reduction to Linear Form

A LevelAQAEdexcelOCR

## 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.

A LevelAQAEdexcelOCR

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

$y=ax^{n}$

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.

A LevelAQAEdexcelOCR

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

$y=ab^{x}$

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.

A LevelAQAEdexcelOCR

@mmerevise

A LevelAQAEdexcelOCR

## 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}

A LevelAQAEdexcelOCR

## 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.

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

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:

$y=4x^{3}$

A LevelAQAEdexcelOCR

## Reduction to Linear Form Example Questions

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 $9.5$

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 $y=729\times0.667^{x}$

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 $2$

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 $y=100\times x^{1.5}$

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

A Level

A Level

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