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I am trying to move the $x$ on the right over to the left side of this equation to solve for $x$:

$$x = \large e^\frac{{{{ \Large(z / x - 1 - 0.2029)}}}} {{\large {-0.022}}}$$

I am basically trying to get the reverse of equations I have to get $z$:

$$y = -0.022 *\ln(x) + 0.2029 + 1$$ $$z = xy$$

I am have created the above equations, but having trouble figuring out how to move X out of the base e's exponent. Any help would be appreciated. Thanks.

fanfavorite
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1 Answers1

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You will not be able to do that: there is no way to bring $x$ "out of the base $e$'s exponent" in the simple equation $x=e^{z/x}$. In fact the solution, which is $x=z/W(z)$, involves the Lambert $W$ function.

Your equation is of the form $$ x = e^{a(z/x+b)}. $$ The solution is $$ x = \frac{az}{W(ae^{-ab}z)} . $$

Federico
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  • Thanks for the help. I don't fully follow how you created that solution and how to apply to mine. a(z/x+b) can convert to az/(ax+ab) I believe and then using W to get out of e exponent, but lost from there. My equation might be easier to work with in this form: x = z / (−0.022 ∗ ln(x) + 0.2029 + 1) – fanfavorite Dec 11 '18 at 19:06
  • "a(z/x+b) can convert to az/(ax+ab) I believe" No! $z/x+b\neq z/(x+b)$! – Federico Dec 12 '18 at 13:05
  • Your equation is of the form I wrote with $a=-1/0.022$ and $b=-1.2029$. – Federico Dec 12 '18 at 13:29