HELP to solve an exponential equation with one unknown

In summary, the conversation revolves around a complex equation that needs to be solved for 'B' in terms of other variables. The equation is in the form of a_1 e^{-b_1 B} + a_2 e^{-b_2 B} + a_3 e^{-b_3 B} = 0 and there is a discussion about the best method to solve it, with suggestions including using Excel Solver or other numerical equation-solving methods. The purpose of solving the equation is to validate the accuracy of a piece of calibrated equipment, as required by an Australian Standard. There is also mention of using open-source software and tutorials for Excel Solver.
  • #1
LanieH
5
0
Hi,
I have an equation with one unknown. This is the 'B' in the equation. I need to solve for 'B'. Can you help me to rearrange this complex equation to solve for B in terms of everything else please?

=(() - (DCRMgAl)) ∗ ^(−∗(()+()))+ ((DCRMgAl )- (DCRAl ))∗ ^(−∗∗() )+((Al) - (DCRMg)) ∗ ^(−∗(()+()))
 
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  • #2
��=((����������) - (DCRMgAl)) ∗ ��^(−��∗((������)+(������)))+ ((DCRMgAl )- (DCRAl ))∗ ��^(−��∗��∗(������) )+((������Al) - (DCRMg)) ∗ ��^(−��∗((������)+(����������)))
... well it's perfectly simple, all you have to do is put � = �_(−��) and then it should all be clear :D
 
  • #3
0=((DCRMg) - (DCRMgAl)) ∗ e^(-B∗((ρMg)+(ρAl)))+ ((DCRMgAl )- (DCRAl ))∗e^(-2∗B∗(ρMg))+((DCRAl) - (DCRMg)) ∗e^(-B∗((ρMg)+(ρMgAl)))

Haha :-) I hope that I can enter the equation correctly this time...
 
  • #4
LanieH said:
0=((DCRMg) - (DCRMgAl)) ∗ e^(-B∗((ρMg)+(ρAl)))+ ((DCRMgAl )- (DCRAl ))∗e^(-2∗B∗(ρMg))+((DCRAl) - (DCRMg)) ∗e^(-B∗((ρMg)+(ρMgAl)))

Haha :-) I hope that I can enter the equation correctly this time...

It looks like you have an equation of the form
[tex] a_1 e^{-b_1 B} + a_2 e^{-b_2 B} + a_3 e^{-b_3 B} = 0,[/tex]
where
[tex] a_1 = DCRMg - DCRMgAl, \, a_2 = DCRMgAl - DCRAl, \, a_3 = DCRAl - DCRMg\\
b_1 = \rho Mg + \rho Al, \, b_2 = 2 \rho Mg, \, b_3 = \rho Mg + \rho MgAl.[/tex]
Is that correct? If so, you had better have mixed signs for a1, a2 and a3, because if they all have the same sign there cannot be any real solution. If they DO have mixed signs, just use a standard numerical equation-solving method, or use a computer package such as Wolfram Alpha, or the EXCEL Solver.

RGV
 
  • #5
Thank you for your help. I have been recently told that this equation must be solved using Excel Solver. I have no idea how to use this program in order to calculate B.
I have created an excel formula relative to the above equation, but I am stuck at this point. Are you able to show me how you would solve the equation for B in Excel Solver?

=(B50-C50)*EXP(-B*($H$44+$I$44))+(C50-D50)*EXP((-2*B($H$44)))+(D50-B50)*EXP((-B($H$44+$J$44)))
 
  • #6
Oh I hate courses like that - it is as if the teacher is a Microsoft sales rep.
(I'm guessing the Mg and Al stuff are subscripts? So ##\rho_{Mg}## would be the density of magnesium?)

FWIW since I too have no experience nor interest in Excel.
How would the marker know you didn't use a different approach to solve it?
Say - Newton-Raphson in gnu-octave or something?
 
  • #7
This is a formula used in an Australian Standard. I need to validate it in Excel Solver to prove that a piece of calibrated equipment is providing me with close to accurate results. Since I cannot solve the equation on paper, I require a program to do it for me. This is where Solver was suggested to me.
 
  • #8
LanieH said:
This is a formula used in an Australian Standard. I need to validate it in Excel Solver to prove that a piece of calibrated equipment is providing me with close to accurate results. Since I cannot solve the equation on paper, I require a program to do it for me. This is where Solver was suggested to me.

If the EXCEL Solver is all you have access to, then go ahead and use it. However, there are much better, much more accurate solvers available. Did you miss my suggestion to use Wolfram Alpha?

Actually, the type of problem you have could easily be done manually, using nothing fancier than a scientific hand-held calculator and Newton's method. Your results obtained in this way might well be more accurate than those from the EXCEL Solver!

RGV
 
  • #9
Yeah, why should the equipment need to be validated with excel in particular and not any other bit of software?
(Maybe the firm has a deal with Microsoft?)

I'm used to doing this the other way around - since - for scientific transparency it is good practice that all parts of the software used in a calculation be open, in principle at least, to scrutiny. That pretty much means open source. So I tend to end up validating results from closed source tools using open source or even just writing my own program and using an open-source compiler.

However - not to be a total killjoy, if solver you must use then solver it is.
There are plenty of references and tutorials online -eg:
http://www.solver.com/excel-solver-help
 
  • #10
LanieH said:
This is a formula used in an Australian Standard. I need to validate it in Excel Solver to prove that a piece of calibrated equipment is providing me with close to accurate results. Since I cannot solve the equation on paper, I require a program to do it for me. This is where Solver was suggested to me.

I don't understand what your requirements are. I understand that you have some equipment that needs to be checked or calibrated. To do that, you need to solve some equation and then do something with the solution. Have I got that correct so far?

At this point you lose me completely: why should it matter what method you use to obtain the solution? Suppose you e-mailed me all the numerical values of the input parameters and I e-mailed you back the numerical solution of the problem (which you could easily check just by substitution). Would that not be all you need? Are you being forced to do it in EXCEL?

RGV
 

FAQ: HELP to solve an exponential equation with one unknown

How do I solve an exponential equation with one unknown?

To solve an exponential equation with one unknown, you will need to use logarithms. First, isolate the exponential term on one side of the equation. Then, take the logarithm of both sides using the same base. This will allow you to bring the exponent down as a coefficient and solve for the unknown variable.

What is the basic process for solving an exponential equation with one unknown?

The basic process for solving an exponential equation with one unknown is to isolate the exponential term, take the logarithm of both sides, bring the exponent down as a coefficient, and solve for the unknown variable.

Can I use any base for the logarithm when solving an exponential equation with one unknown?

Yes, you can use any base for the logarithm when solving an exponential equation with one unknown. However, it is recommended to use the base of the exponential term to simplify the equation.

What if the exponential term is on both sides of the equation?

If the exponential term is on both sides of the equation, you will need to use the properties of logarithms to simplify the equation. One property states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. Another property states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and denominator.

Are there any special cases when solving an exponential equation with one unknown?

Yes, there are two special cases when solving an exponential equation with one unknown. The first is when the base of the exponential term is equal to 1. In this case, the solution will always be 0. The second is when the base is equal to 0. In this case, there will be no solution.

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