Maximum value equation: a constant to the power of x

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SUMMARY

The discussion focuses on solving the equation L = c^3.(0.5)^(c-1).(0.6)^(c-1).(0.8)^(c-1).(1 - (0.5)^c)^42.(0.5)^(8c) for the variable c. The user attempts to find the maximum value by taking the natural logarithm of both sides and differentiating, ultimately leading to a complex equation involving the Lambert W function. The conclusion is that the equation does not have a solution in terms of elementary functions, necessitating the use of advanced mathematical tools like the Lambert W function for further analysis.

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  • Understanding of natural logarithms and differentiation
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Homework Statement


Hi everyone, this is part of a longer question, but the current part involves me finding the value of c in the following equation:

L = c^3.(0.5)^(c-1).(0.6)^(c-1).(0.8)^(c-1).(1- (0.5)^c)^42.(0.5)^(8c)

Homework Equations





The Attempt at a Solution


Here's what I've done so far:

I take the natural log of both sides:

ln L = 3.ln(c) + (c-1)ln(0.5) + (c-1)ln(0.6) + (c-1)ln(0.8) + 42ln(0.5^c - 1) + 8c.ln(0.5)

Then differentiate to find the max:

∂(lnL)/∂c = 3/c + ln(0.5) + ln(0.6) + ln(0.8) + [42[ln(0.5)](0.5)^c]/(0.5^c - 1) + 8ln(0.5)

At max, ∂(lnL)/∂c = 0

Relabel [ln(0.5) + ln(0.6) + ln(0.8) + 8ln(0.5)] as (-A) for convenience

Therefore,
3/c + [42[ln(0.5)](0.5)^c]/(0.5^c - 1) = A

Multiplying across by c(0.5^c - 1) gives:

3(0.5^c) - 3 + 42.ln(0.5)c(0.5)^c = Ac(0.5)^c - Ac

Relabel 42.ln(0.5) as B for convenience

Rearranging, we get:

(3 + Bc - Ac)(0.5)^c = 3 - Ac

c.ln(0.5) = ln ((3-Ac)/(3 + Bc + Ac))

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And this is where I run into problems, as I can't get rid of the ln on the right side without making an e^c on the left, so I don't know how to isolate the c. I tried differentiating both sides, but I'm pretty sure you can't do that. Is my entire method wrong? The equation at the start is definitely correct though. Please could someone point me in the right direction?

Thanks!
 
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An equation like that, with the unknown value both in, and outside of, a transcendental function, here the exponential, is not going to have a solution in terms of an elementary function. You might be able to change it so that you can use "Lambert's W function", which is defined as the inverse function to f(x)= xe^x
 

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