1. ## Solving an equation.

I'm working on implementing an old method for determination of cerebral blood flow in rats. I'm stuck in this equation! It is some years since I've worked with such formulas, and I can not get it to go up.

I really hope that there is one who will try .. X supposed to be equal to 0.058 / sec

(A/B)=C* ((exp[X-T*0,32])-1)/X

A=0,0026 nCi (Curie, measurement of radio-activity)
B=0,00152 mL/second (Millilitre/sec)
C=0,074 uCi/mL
T=20 seconds

This is the actual equation(11) from the article with values and constants ..:

Here, x is called k..

Thanks/ Jakob

2. Originally Posted by jakobbokaj
I'm working on implementing an old method for determination of cerebral blood flow in rats. I'm stuck in this equation! It is some years since I've worked with such formulas, and I can not get it to go up.

I really hope that there is one who will try .. X supposed to be equal to 0.058 / sec

(A/B)=C* ((exp[X-T*0,32])-1)/X

A=0,0026 nCi (Curie, measurement of radio-activity)
B=0,00152 mL/second (Millilitre/sec)
C=0,074 uCi/mL
T=20 seconds

This is the actual equation(11) from the article with values and constants ..:

Here, x is called k..

Thanks/ Jakob
You cannot solve for X algebraically. A numerical procedure has to be used to get a decimal approximation. Use a graphics or CAS calculator or computer software to solve it.

3. I've used my 'TexasInstruments TI83' as you describe - and my problem is more accurately that I can not get the same value for X (or k) as the author. I've tried hard and suspect an error in the article.

Is there anyone who can get the right value of X?

4. Originally Posted by jakobbokaj
I've used my 'TexasInstruments TI83' as you describe - and my problem is more accurately that I can not get the same value for X (or k) as the author. I've tried hard and suspect an error in the article.

Is there anyone who can get the right value of X?
solve 0.0026&#47;0.00152 &#61; 0.074&#40;Exp&#91;x - 2&#40;0.32&#41;&#93; - 1&#41;&#47;x - Wolfram|Alpha

5. I'm seeing some mixed units in there. A and C, for example, have different units. Have you taken that into account?

6. Originally Posted by Ackbeet
I'm seeing some mixed units in there. A and C, for example, have different units. Have you taken that into account?
Good call.

The OP can edit the input at the Wolfram Alpha link I posted.

7. There's something else very strange with that equation. The expression

$e^{k-Tkt'}$

is highly problematic. It appears that the units of T are time, as well as t'. The units of k are inverse time. But the argument of an exponential function is not allowed to have units of any sort: it must be dimensionless. Otherwise, in the Taylor series expansion, you'd have widely varying powers of units being added together.

Might there be an error in the formula?

8. Maybe the correct argument is $kT-kt'=k(T-t')?$ That would work in terms of units.

9. Great point. I really appreciate your responses. Here you can see how they reach equation (11) .. They want to bring k and t' together lige that because the value of kt' is known.

10. Originally Posted by Ackbeet
Maybe the correct argument is $kT-kt'=k(T-t')?$ That would work in terms of units.
YES - that way I get the right solution. I am very happy and relieved.

http://www.wolframalpha.com/input/?i...%5D+-+1%29%2Fx

- Have a good time -

11. Good. My guess about the argument is correct. If you compute the integral and simplify, you get exactly equation 11, but with my argument for the exponential function. Therefore, I conclude that equation 11 contains a typo. Obviously, the authors simply mis-typed it, since they got the correct number for k. That is, they used my argument, not the one they typed up, when they did their computations.

Cheers.