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kp^(1-a)-bp^(-a)-c=0

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Shashank kumar 2018-7-12

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关闭： MATLAB Answer Bot 2021-8-20

I want to write the above equation for P. Is it possible to do that in Matlab?

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Walter Roberson 2018-7-12

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It could certainly be solved numerically if you have the other values.

If you need to solve symbolically then I predict that there will be problems. I am away from my desktop at the moment so I cannot check, but my visual examination suggests that either there is no closed form solution, or else that it just might have a solution involving LambertW. Unfortunately MATLAB is weak on solutions involving LambertW.

If no one else answers then I will check my tools when I get back to my desktop in a few hours.

Walter Roberson 2018-7-12

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There does not appear to be a closed form solution for this. You would need a numeric approach.

Stephan 2018-7-12

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编辑：Stephan 2018-7-12

solve_for_p.mlx

Perhaps it is a consolation to simplify the equation a bit and eliminate the power p ^ (1-a) - (unless I made a mistake). This step may speed up the numerical calculation of p.

Walter Roberson 2018-7-12

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在 MATLAB Online 中打开

Stephan,

You do

p = isolate(f,p)

which assigns to p an equation involving p. If you are replacing p, it should be with an expression not an equation.

Then when you work with lhs(f) and rhs(f) on the next line, you do not do a subs() to bring in the revised value of p, so everything gets confused at that point.

Stephan 2018-7-12

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编辑：Stephan 2018-7-12

在 MATLAB Online 中打开

I corrected already with a new attachment... result keeps the same. m-file would look like:

syms a b c p k
f = k*p^(1-a) - b*p^(-a) - c == 0
f = f + c + b*p^(-a)
f_lhs = expand(log(lhs(f)),'IgnoreAnalyticConstraints',true)
f_rhs = expand(log(rhs(f)) ,'IgnoreAnalyticConstraints',true)
f = f_lhs == f_rhs
f = simplify(f)
f = exp(lhs(f)) == exp(rhs(f))
f = simplify(f)

which results in:

f =
b + c*p^a == k*p

Walter Roberson 2018-7-12

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Okay, I pick up the corrected version now.

You can get the same thing more directly by multiplying all the terms by p^a

Stephan 2018-7-12

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Nice - i didnt see that it is that easy...

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k*p^(1-a)-b*p^(-a)-c=0

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kp^(1-a)-bp^(-a)-c=0

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