Using logs to obtain the slope of a function

[thepassenger48]

Homework Statement

I have to find the slope of an exponential curve by using log identities.

Using basic values such as:
(10, 0.692)
(30, 1.082)
(70, 1.579)

Homework Equations

The equation is in the form T=Cm^p
So the log function is logT=logC + Plog(m) right?

The Attempt at a Solution

When I take the log of (10, 0.692) I end up with negative values, which is completely non-sensical when compared to my experimental values!
I end up with function -.6x^.4
when I should get .3x^.4

 

[Kreizhn]

To me there is still a fair bit of ambiguity in the statement of your problem, perhaps if you could elaborate a bit further, it would be easier to assist you. For example, the ordered pair (10, 0.692), what values does this correspond to in the equation? Your formula is certainly right, but I'm not quite following. I assume the abscissa is Temperature, but what's the other?

 

[HallsofIvy]

Homework Statement

I have to find the slope of an exponential curve by using log identities.

Using basic values such as:
(10, 0.692)
(30, 1.082)
(70, 1.579)

Homework Equations

The equation is in the form T=Cm^p
So the log function is logT=logC + Plog(m) right?

The Attempt at a Solution

When I take the log of (10, 0.692) I end up with negative values, which is completely non-sensical when compared to my experimental values!
I end up with function -.6x^.4
when I should get .3x^.4

Why is it non-sensical? you are talking about the logarithm being negative, not the values themselves.

I'm going to assume that the first number in each pair is m, the second number T. As Kreizhn said, it would be good idea to actually put that in your question.

(log 10, log 0.692)= (1, -0.160)
(log 30, log 1.082)= (1.477, 0.0342)
(log 70, log 1.579)= (1.845, 0.198)

slope, P:
(0.198-(-1.60))/(1.845- 1)= 0.358/0.845= 0.424
When m= 10, log 0.692= -.160= log C+ (0.424)log 10= log C+ 0.424 so
log C= -.160- 0.424= -.584 which means C itslelf is less than 1 but positive:
log T= -.584+ 0.424log m so T= 10^-.584 m^0.424= 0.2550m^0.424. Rounded to one decimal place, that is T= .3 m^.4.

log C is -.6, C= .3 (again, to one decimal place)

 

[thepassenger48]

Why is it non-sensical? you are talking about the logarithm being negative, not the values themselves.

I'm going to assume that the first number in each pair is m, the second number T. As Kreizhn said, it would be good idea to actually put that in your question.

(log 10, log 0.692)= (1, -0.160)
(log 30, log 1.082)= (1.477, 0.0342)
(log 70, log 1.579)= (1.845, 0.198)

slope, P:
(0.198-(-1.60))/(1.845- 1)= 0.358/0.845= 0.424
When m= 10, log 0.692= -.160= log C+ (0.424)log 10= log C+ 0.424 so
log C= -.160- 0.424= -.584 which means C itslelf is less than 1 but positive:
log T= -.584+ 0.424log m so T= 10^-.584 m^0.424= 0.2550m^0.424. Rounded to one decimal place, that is T= .3 m^.4.

log C is -.6, C= .3 (again, to one decimal place)

That was a flawless answer. Thank you very much.
I'm sorry for wording my question so poorly.