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How to solve for $x$ : $ae^{bx}+x = c$

I've tried to solve it with the Lambert W function, but, in all of the methods I tried, I had $x$ as the Lambert W function parameter. Can I solve it without graphs? Thank you for your help.

Gabriel Augusto
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  • This might prove useful to solve the related

    $$(e^b)^x - \left( - \frac 1 a \right) x = \frac c a$$

    which after multiplication by $a$ should give you your equation.

     – PrincessEev Dec 29 '20 at 20:52
  • WolframAlpha shows a solution here as

    $$x = c - \frac{W_n(a b e^{b c})}{b}\quad \text{for } b\ne0 \land a b\ne0 \land n \in\mathbb{ Z}$$

     – poetasis Dec 29 '20 at 20:54

1 Answers1

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$$ae^{bx}+x=c$$

$$ae^{bx}=c-x$$

$$ae^{bc-b(c-x)}=c-x$$

$$abe^{bc}=b(c-x)e^{b(c-x)}$$

$$b(c-x)=W(abe^{bc}).$$

$$x=c-\frac1bW(abe^{bc}).$$

(By hand.)