For instance, how would I solve: $3^x + x = 85$ ?
How to solve exponential function of form $a b^x + x = c$?
-
0Or perhaps you meant $b^x+x=c$? – 2012-11-05
3 Answers
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.
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 $x<0$
$a b^x + x = c$
$a b^x = c-x$
$ 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 $a b^x + x = c$
For your example:
$3^x + x = 85$
$x=85-\frac{W(3^{85}\ln{85})}{\ln{85}}$