MathCADÆs solve for variable in the symbolic menu

[MURZ]

I am currently working on a project where I am using MathCAD’s solve for variable in the symbolic menu. I have the following

First equation is ?m := 1/n3 where n3 is 11.408 this implies that ?m = 0.0877

I then try to solve for ? in the following equation using the solve for variable in the symbolic menu.
?m = (1/?)exp(-?-?) (For this equal sign I used Ctrl =) when I use the solve for variable I get ?=1/2ln(2/?m) which implies ?=1.564.

However, when I set the equation like this 0.0877 = (1/?)exp(-?-?) and solve for ? I get 1.14793 which is the answer I need.

Why are the two different? How can I get the equation to equal the 1.1479 when I use the variable ?m in the equation. I would eventually like to use the equation

Acm:=(1/?)exp(-?-?)=?m solve,?
Any help would be appreciated

Thanks

Mark M

 

[jghrist]

I'm confused by the use of the question mark. Is this a variable name? If you type ?, you get a derivative operator.

 

[MURZ]

Let's try this. Looks like when I did a cut and past something went wrong.

I have a mathcad sheet if it will help.

I am currently working on a project where I am using MathCAD’s solve for variable in the symbolic menu. I have the following

First equation is Em := 1/n3 where n3 is 11.408 this implies that Em = 0.0877

I then try to solve for E in the following equation using the solve for variable in the symbolic menu.
Em = (1/E)exp(-E-E) (For this equal sign I used Ctrl =) when I use the solve for variable I get E=1/2ln(2/Em) which implies E=1.564.

However, when I set the equation like this 0.0877 = (1/E)exp(-E-E) and solve for E I get 1.14793 which is the answer I need.

Why are the two different? How can I get the equation to equal the 1.1479 when I use the variable ?m in the equation. I would eventually like to use the equation

Acm:=(1/E)exp(-E-E)=Em solve,E
Any help would be appreciated

 

[GregLocock]

so, if you back substitute E=1/2ln(2/Em) into (1/E)exp(-E-E)
does it make sense?

have you tried plotting the function out to see why there might be a problem?



Cheers

Greg Locock

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