How to isolate a variable in two terms of the equation exponencial

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roberto2020

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Hi everyone, I'm trying to isolate de variable "D", but I'm not having success. Could someone help me out?



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Can't be done. Anything else? How much time have you spent trying to do it?
 
roberto2020 said:
I've spent a lot of time.

Why does it can't be solved?
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It is a common disappointment when algebra students learn that they can't accomplish what is wanted with wonderfully simple things like [math]x=e^x[/math].

All is not lost. There are ways to find specific solutions with specific values.
 
Reverse situation is that when you learn that you need no more than 4 colors to color maps.....
 
Whether this can or cannot be solved depends on what you are wiling to accept as answer- what functions you have available.

I would start by focusing on what is important. Since D appears in the denominator of two different fractions, let x= 1/D. The equation can be written as \(\displaystyle (BCx- BEF)e^{-Hx}= A\). Let y= BCx- BEF so x= (y- BEF)/BC and the equation is \(\displaystyle ye^{-Hy}e^{HEF/C}= a\) and then \(\displaystyle ye^{-Hy}= Ae^{HEF/C}\). Finally, let z= -Hy so that \(\displaystyle -ze^z/H= Ae^{HEF/C}\) so \(\displaystyle ze^z= -AH e^{HEF/C}\).

Now we turn to "Lambert's W function", W(x), also called "omega function" or "log product function". It is defined as the inverse function to \(\displaystyle f(x)= xe^x\). Using that, we get \(\displaystyle z= W(-AH e^{HEF/C}\).

Now go back. z= -Hy so \(\displaystyle y= -z/H= -W(-AHe^{HEF/C})/H\). \(\displaystyle y= BCx- BEF\) so \(\displaystyle BCX- BEF= -W(-AHe^{HEF/C})/H. BCx= BEF- W(-AHe^{HEF/C})/H\). \(\displaystyle x= EFC- W(-AHe^{HEF/C})/BCH\). Finally, x= 1/D so D= 1/x.
 
Professor Halls, this problem is listed under arithmetic.

To OP, usually if D is in both the base and exponent you can't solve for it. Sometimes you will be able to see a solution (as in x^x =4, where x=2 works) but other times you can't solve the equation.
 
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