Reduction to Linear Form

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Reduction to Linear Form

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

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y=axnmathbf{y=ax^{n}}

y=axny=ax^{n}

Take logarithms:

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

Overall we have:

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

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

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

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y=abxmathbf{y=ab^{x}}

y=abxy=ab^{x}

Take logarithms:

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

Overall we have:

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

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

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

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Example 1: Converting to Linear Form

Convert y=99×0.5xy=99times0.5^{x} to linear form.

[2 marks]

y=99×0.5xy=99times0.5^{x}

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

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Example 2: Using Linear Form

The data below is taken from a plot of the form y=axny=ax^{n}. By plotting log10(y)log_{10}(y) against log10(x)log_{10}(x), find aa and nn.

[5 marks]

Step 1: Calculate log10(x)log_{10}(x) and log10ylog_{10}y and put them in a table.

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

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

Gradient is approximately 3.30.60.90=2.70.9=3dfrac{3.3-0.6}{0.9-0}=dfrac{2.7}{0.9}=3

y-intercept is approximately 0.60.6

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

y=axnlog(y)=log(a)+nlog(x)y=ax^{n}rightarrow log(y)=log(a)+nlog(x)

Gradient is nn so n=3n=3

y-intercept is log(a)log(a) so log(a)=0.6log(a)=0.6 so a=100.6=4a=10^{0.6}=4 to two significant figures.

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

y=4x3y=4x^{3}

 

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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=abxy=ab^{x} is monitored over a number of years. Use the data from the monitoring to find aa and bb.

 

 

[5 marks]

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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 22.

 

Note the linear form:

y=abxlog(y)=xlog(b)+log(a)y=ab^{x}rightarrow log(y)=xlog(b)+log(a)

So the straight line is obtained by plotting log(y)log(y) against xx.

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

 

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

 

 

Find the gradient and the y-intercept.

Gradient is 69.560=3.56=712dfrac{6-9.5}{6-0}=dfrac{3.5}{6}=-dfrac{7}{12}

y-intercept is 9.59.5

 

Interpret the gradient and y-intercept.

Gradient is log(b)log(b)

log(b)=712log(b)=-dfrac{7}{12}

b=0.667b=0.667

 

y-intercept is log(a)log(a)

log(a)=9.5log(a)=9.5

a=729a=729

 

Put it all together:

Our estimate for the line is y=729×0.667xy=729times0.667^{x}

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

[5 marks]

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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 1010.

 

Note the linear form:

y=axnlog(y)=nlog(x)+log(a)y=ax^{n}rightarrow log(y)=nlog(x)+log(a)

So the straight line is obtained by plotting log(y)log(y) against log(x)log(x).

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

 

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

 

 

Find the gradient and the y-intercept.

Gradient is 3.5210=1.51=1.5dfrac{3.5-2}{1-0}=dfrac{1.5}{1}=1.5

y-intercept is 22

 

Interpret the gradient and y-intercept.

Gradient is nn

n=1.5n=1.5

 

y-intercept is log(a)log(a)

log(a)=2log(a)=2

a=100a=100

 

Put it all together:

Our estimate for the line is y=100×x1.5y=100times x^{1.5}

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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=abxy=ab^{x} curve. Use the table of census data to find aa and bb.

[5 marks]

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Again we can choose our own base for logarithms, so we will take the logarithm with base 33.

 

y=abxy=ab^{x}

 

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

 

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

 

 

 

Plotting log(y)log(y) against xx gives this graph.

 

 

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

 

gradient=1012.357101=2.3579=0.262begin{aligned}text{gradient}&=dfrac{10-12.357}{10-1}[1.2em]&=dfrac{-2.357}{9}[1.2em]&=-0.262end{aligned}

 

log(b)=0.262log(b)=-0.262

 

b=30.262b=3^{-0.262}

 

b=0.750b=0.750

 

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

 

So yy intercept is 12.61912.619

 

log(a)=12.619log(a)=12.619

 

a=312.619a=3^{12.619}

 

a=1048576a=1048576

 

Hence, y=1048576×0.750xy=1048576times0.750^{x}

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Specification Points Covered

F6 – Use logarithmic graphs to estimate parameters in relationships of the form y=axny=ax^n and y=kbxy=kb^x, given data for xx and yy