B Making "c" the Subject of a=bc(d)^c

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Making "c" the subject of the equation a = bc(d)^c is challenging due to its complexity. The discussion reveals that traditional methods, such as logarithms, do not yield a solution. It is suggested that the Lambert W function is necessary for solving this equation, as it cannot be expressed using standard algebraic functions. The terms "b" and "d" are constants, which further complicates isolating "c." Ultimately, the consensus is that the equation is not solvable using elementary algebra.
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Apologises if this question is not appropriate for this thread but it seemed less suitable to the homework thread as I'm not sure whether it's possible (although I'm hopeful that it is). This question is a product of my self- study of maths and that I've already asked my maths teacher; who was unable to answer.

The question is this: how do I make the term "c" the subject of the equation below?

a = bc (d)^c

Many thanks for your time.
 
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Einstein's Cat said:
Apologises if this question is not appropriate for this thread but it seemed less suitable to the homework thread as I'm not sure whether it's possible (although I'm hopeful that it is). This question is a product of my self- study of maths and that I've already asked my maths teacher; who was unable to answer.

The question is this: how do I make the term "c" the subject of the equation below?

a = bc (d)^c

Many thanks for your time.

I was not able to solve this using logarithms. Probably, this is not solvable using elementary algebra.
 
Math_QED said:
I was not able to solve this using logarithms. Probably, this is not solvable using elementary algebra.
Are there any alternative methods in which it could be solved?

Also, I must say that the terms "d" and "b" are constant.
 
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What you ask is impossible. It can be proven to be impossible. You need what is called the Lambert W function for this. This function does not have a repesentation in terms of the standard functions. https://en.wikipedia.org/wiki/Lambert_W_function
 
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