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

[roberto2020]

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

 

[tkhunny]

Can't be done. Anything else? How much time have you spent trying to do it?

 

[roberto2020]

Can't be done. Anything else? How much time have you spent trying to do it?

I've spent a lot of time.

Why does it can't be solved?

 

[Deleted member 4993]

I've spent a lot of time. Why does it can't be solved?

That is the nature of the beast!

 

[tkhunny]

I've spent a lot of time.

Why does it can't be solved?

It is a common disappointment when algebra students learn that they can't accomplish what is wanted with wonderfully simple things like $$x=e^x$$.

All is not lost. There are ways to find specific solutions with specific values.

 

[Deleted member 4993]

Reverse situation is that when you learn that you need no more than 4 colors to color maps.....

 

[HallsofIvy]

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.

 

[Steven G]

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.

 

[HallsofIvy]

I wouldn't consider a problem with exponentials to be "arithmetic"!