# Reduction of Equations of the form y=ax^(n) to linear form?

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Hi,

The question is as follows:

(Two quantities Q and H are believed to be related by the equation Q=kH^(n). The experimental values obtained for Q and H are shown below:

Q = 0.16 0.20 0.27 0.34 0.40 0.47 0.55

H = 1.14 1.78 3.24 5.14 7.11 9.82 13.44

Determine the law connecting Q and H, showing on your graph the point of interception and the gradient.)

Ok so I understand so far that I have to use the laws of logs to reduce a "law" of the type y=ax^(n) to straight line form and after this using, log graph paper, i gotta plot the graph and get the values for constants a and n. However, I kinda dont know where to start. In my notes it says i need to "take logs to both sides of the equation so that it becomes log y = log (ax)^(N) and log y = log a + log x^(n)" but thats as far as my understanding goes. If anyone could guide me in the right direction for this question and help my understanding that would be great. Thanks alot.

The question is as follows:

(Two quantities Q and H are believed to be related by the equation Q=kH^(n). The experimental values obtained for Q and H are shown below:

Q = 0.16 0.20 0.27 0.34 0.40 0.47 0.55

H = 1.14 1.78 3.24 5.14 7.11 9.82 13.44

Determine the law connecting Q and H, showing on your graph the point of interception and the gradient.)

Ok so I understand so far that I have to use the laws of logs to reduce a "law" of the type y=ax^(n) to straight line form and after this using, log graph paper, i gotta plot the graph and get the values for constants a and n. However, I kinda dont know where to start. In my notes it says i need to "take logs to both sides of the equation so that it becomes log y = log (ax)^(N) and log y = log a + log x^(n)" but thats as far as my understanding goes. If anyone could guide me in the right direction for this question and help my understanding that would be great. Thanks alot.

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#2

(Original post by

Hi,

The question is as follows:

(Two quantities Q and H are believed to be related by the equation Q=kH^(n). The experimental values obtained for Q and H are shown below:

Q = 0.16 0.20 0.27 0.34 0.40 0.47 0.55

H = 1.14 1.78 3.24 5.14 7.11 9.82 13.44

Determine the law connecting Q and H, showing on your graph the point of interception and the gradient.)

Ok so I understand so far that I have to use the laws of logs to reduce a "law" of the type y=ax^(n) to straight line form and after this using, log graph paper, i gotta plot the graph and get the values for constants a and n. However, I kinda dont know where to start. In my notes it says i need to "take logs to both sides of the equation so that it becomes log y = log (ax)^(N) and log y = log a + log x^(n)" but thats as far as my understanding goes. If anyone could guide me in the right direction for this question and help my understanding that would be great. Thanks alot.

**dc2209**)Hi,

The question is as follows:

(Two quantities Q and H are believed to be related by the equation Q=kH^(n). The experimental values obtained for Q and H are shown below:

Q = 0.16 0.20 0.27 0.34 0.40 0.47 0.55

H = 1.14 1.78 3.24 5.14 7.11 9.82 13.44

Determine the law connecting Q and H, showing on your graph the point of interception and the gradient.)

Ok so I understand so far that I have to use the laws of logs to reduce a "law" of the type y=ax^(n) to straight line form and after this using, log graph paper, i gotta plot the graph and get the values for constants a and n. However, I kinda dont know where to start. In my notes it says i need to "take logs to both sides of the equation so that it becomes log y = log (ax)^(N) and log y = log a + log x^(n)" but thats as far as my understanding goes. If anyone could guide me in the right direction for this question and help my understanding that would be great. Thanks alot.

**log(a^b)=b*log(a)**and then you have a linear equation which you plot on the graph.

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(Original post by

Your log law is right so far but you should now know that

**Aph**)Your log law is right so far but you should now know that

**log(a^b)=b*log(a)**and then you have a linear equation which you plot on the graph.How is that a linear equation?

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#4

(Original post by

How is that a linear equation?

**dc2209**)How is that a linear equation?

**y=mx+c**

you have

**log(y)=n*log(x)+log(a)**

therefore n=m and log(a)=c

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#5

(Original post by

How is that a linear equation?

**dc2209**)How is that a linear equation?

n and a are constants and you want something that looks like y=mx+c

If your axes are log(y) and log(x) instead of y and x then you have a linear equation

Where n is the gradient and log(a) is your intercept

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(Original post by

n and a are constants and you want something that looks like y=mx+c

If your axes are log(y) and log(x) instead of y and x then you have a linear equation

Where n is the gradient and log(a) is your intercept

**TenOfThem**)n and a are constants and you want something that looks like y=mx+c

If your axes are log(y) and log(x) instead of y and x then you have a linear equation

Where n is the gradient and log(a) is your intercept

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#7

(Original post by

Ah, I understand. So I have to plot this on logarithmic graph paper?

**dc2209**)Ah, I understand. So I have to plot this on logarithmic graph paper?

You do not need to - if you plot log(x) against log(y) - ie you take logs and plot those values - it will still work

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(Original post by

you can

You do not need to - if you plot log(x) against log(y) - ie you take logs and plot those values - it will still work

**TenOfThem**)you can

You do not need to - if you plot log(x) against log(y) - ie you take logs and plot those values - it will still work

So as far as plotting goes, I have to take the values of Q and H and put those into the logarithmic equation?

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#9

(Original post by

So as far as plotting goes, I have to take the values of Q and H and put those into the logarithmic equation?

**dc2209**)So as far as plotting goes, I have to take the values of Q and H and put those into the logarithmic equation?

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(Original post by

yes

**TenOfThem**)yes

The equation log (y) = nlog(x) + log(a)?

And I replace the "x" in the "nlog (x)" part with every single value of both Q and H to obtain a value for "log (y)"?

if so what would the value of "log (a)" be?

Thanks

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#11

(Original post by

The equation log (y) = nlog(x) + log(a)?

And I replace the "x" in the "nlog (x)" part with every single value of both Q and H to obtain a value for "log (y)"?

if so what would the value of "log (a)" be?

Thanks

**dc2209**)The equation log (y) = nlog(x) + log(a)?

And I replace the "x" in the "nlog (x)" part with every single value of both Q and H to obtain a value for "log (y)"?

if so what would the value of "log (a)" be?

Thanks

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(Original post by

When you draw the straight line the y-intercept will be log(a)

**TenOfThem**)When you draw the straight line the y-intercept will be log(a)

Ok, so to say, find the value of log (y) when Q = 0.16, I would just put log (0.16) into my calculator?

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#13

(Original post by

Ok, so to say, find the value of log (y) when Q = 0.16, I would just put log (0.16) into my calculator?

**dc2209**)Ok, so to say, find the value of log (y) when Q = 0.16, I would just put log (0.16) into my calculator?

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(Original post by

yes

**TenOfThem**)yes

Ive plotted the graphs- they are just about straight lines I guess.

As for obtaining the values for the constants "a" and "n"...? How would I go about this?

thnknas

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#15

(Original post by

Ive plotted the graphs- they are just about straight lines I guess.

As for obtaining the values for the constants "a" and "n"...? How would I go about this?

thnknas

**dc2209**)Ive plotted the graphs- they are just about straight lines I guess.

As for obtaining the values for the constants "a" and "n"...? How would I go about this?

thnknas

log(a) is the intercept

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Since the log x scales are so different for the two lines, I've had to do them on different graphs, is this ok?

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#17

(Original post by

Since the log x scales are so different for the two lines, I've had to do them on different graphs, is this ok?

**dc2209**)Since the log x scales are so different for the two lines, I've had to do them on different graphs, is this ok?

The y axis was Log(Q) and the x axis was log(H)

So the first point to plot was (log1.14, log0.16) did you plot (0.0569, -0.796) or not

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(Original post by

I am not sure what you have done there is only one line

The y axis was Log(Q) and the x axis was log(H)

So the first point to plot was (log1.14, log0.16) did you plot (0.0569, -0.796) or not

**TenOfThem**)I am not sure what you have done there is only one line

The y axis was Log(Q) and the x axis was log(H)

So the first point to plot was (log1.14, log0.16) did you plot (0.0569, -0.796) or not

Ah, that explains it, I've plotted 0.16 against -0.769 so on

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#19

(Original post by

Ah, that explains it, I've plotted 0.16 against -0.769 so on

**dc2209**)Ah, that explains it, I've plotted 0.16 against -0.769 so on

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