# How to solve exponential function of form $a b^x + x = c$?

For instance, how would I solve: $3^x + x = 85$ ?

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In the title, do you mean $ab^x+x=c$? – Daryl Nov 5 '12 at 5:53
Or perhaps you meant $b^x+x=c$? – Cameron Buie Nov 5 '12 at 6:16

You can solve this using the product log function, its a special function, so if your not used to using them, you might just want a numeric answer. But your special case has the simple solution 4. Here is a link, http://mathworld.wolfram.com/LambertW-Function.html.

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If we consider $x\ge 0,3^x\le 85,x\le 4$

If $x<4,3^x<81\implies 3^x+x<85$

So $x$ can be $4$ which actually satisfies the given equation.

If $x<0,3^x<1\implies -3^x>-1\implies x=85+(-3^x)>84$ which is impossible as $x<0$

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$$a b^x + x = c$$

$$a b^x = c-x$$

$$b^x = \frac{c-x}{a}$$

$$e^{x\ln{b}} = \frac{c-x}{a}$$

$$1= \frac{c-x}{a} e^{-x\ln{b}}$$

$$e^{c\ln{b}}= \frac{c-x}{a} e^{-x\ln{b}} e^{c\ln{b}}$$

$$ae^{c\ln{b}}= (c-x) e^{(c-x)\ln{b}}$$ $$\ln{b}.a.e^{c\ln{b}}= \ln{b}(c-x) e^{(c-x)\ln{b}}$$ $$b^c\ln{b}.a= \ln{b}(c-x) e^{(c-x)\ln{b}}$$

$u=\ln{b}(c-x)$

$$ue^u= b^c\ln{b}.a$$

$$u= W(b^c\ln{b}.a)$$ where $W(x)$ is Lambert W function

$$u=\ln{b}(c-x)=W(b^c\ln{b}.a)$$

$$x=c-\frac{W(a b^c\ln{b})}{\ln{b}}$$

It is general solution of $a b^x + x = c$

$3^x + x = 85$
$$x=85-\frac{W(3^{85}\ln{85})}{\ln{85}}$$