Reduction to Linear Form

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 Level AQA Edexcel OCR

\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
A Level AQA Edexcel OCR

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

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

 

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}

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}

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}

Additional Resources

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