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

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Thread starter 5 years ago
#1
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.
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5 years ago
#2
(Original post by 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.
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.
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Thread starter 5 years ago
#3
(Original post by 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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5 years ago
#4
(Original post by dc2209)
How is that a linear equation?
Liner 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 years ago
#5
(Original post by 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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Thread starter 5 years ago
#6
(Original post by 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
Ah, I understand. So I have to plot this on logarithmic graph paper?
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5 years ago
#7
(Original post by dc2209)
Ah, I understand. So I have to plot this on logarithmic graph paper?
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
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Thread starter 5 years ago
#8
(Original post by 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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5 years ago
#9
(Original post by dc2209)
So as far as plotting goes, I have to take the values of Q and H and put those into the logarithmic equation?
yes
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Thread starter 5 years ago
#10
(Original post by 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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5 years ago
#11
(Original post by 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
When you draw the straight line the y-intercept will be log(a)
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Thread starter 5 years ago
#12
(Original post by 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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5 years ago
#13
(Original post by 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?
yes
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Thread starter 5 years ago
#14
(Original post by 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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5 years ago
#15
(Original post by 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
n is the gradient
log(a) is the intercept
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Thread starter 5 years ago
#16
(Original post by TenOfThem)
n is the gradient
log(a) is the intercept

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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5 years ago
#17
(Original post by 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?
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
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Thread starter 5 years ago
#18
(Original post by 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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5 years ago
#19
(Original post by dc2209)
Ah, that explains it, I've plotted 0.16 against -0.769 so on
One more go then
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5 years ago
#20
(Original post by TheFrobins)
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