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$$e^{A+B*\ln(x)}=m*e^{C+D*\ln(x)}+n,\qquad x \gt 0$$

I'm trying to solve for x in the equation above, but I am not sure where to even begin. If I take the natural log of both sides, I wind up with

$$A+B*\ln(x)=\ln(m*e^{C+D*\ln(x)}+n)$$

which is no nearer to solving it, and I don't know what to do next. I don't know of any natural log properties that would be applicable.

Is this equation possible to solve for x? If so, what's the best way to approach this?

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  • $\begingroup$ Use exponent properties: $$e^{A+B}=e^Ae^B\\(e^A)^B=e^{AB}\\e^{\ln(x)}=x$$definition of logarithm on the last one. $\endgroup$ – Simply Beautiful Art Nov 10 '16 at 21:38
  • $\begingroup$ Is the +$n$ part of the exponent or is it not? By the way you have written, it looks as though it is not part of the exponent. $\endgroup$ – Clayton Nov 10 '16 at 21:39
  • $\begingroup$ Do you have specific values for $B$ or $D$? Or do you know if they are integers? $\endgroup$ – Clayton Nov 10 '16 at 21:43
  • $\begingroup$ A, B, C, D, n, and m are all known and are not integers. n is not part of the exponent which is the main issue when trying to solve. $\endgroup$ – chunky Nov 10 '16 at 22:05
  • $\begingroup$ Thanks @SimpleArt. I feel pretty dumb that I didn't think of that.. got hung up on the logs. $\endgroup$ – chunky Nov 10 '16 at 22:16
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Note that: $$e^{A + B\ln(x)} = e^Ae^{\ln(x^B)} =e^Ax^B$$ Applying that everywhere, we can simplify down to: $$e^Ax^B = me^Cx^D + n$$ As we see, we have a polynomial of degree B or D, whichever is higher. A solution to this depends on the value of B, D and n. There is no elementary solution for any B, D and n.

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  • $\begingroup$ I think at the end you meant for general $B$, $D$, and $n$. As for small enough $B$ and $D$, we have possible solutions. $\endgroup$ – Clayton Nov 10 '16 at 21:45
  • $\begingroup$ Yeah. Replace "any" with "all" or "general". Said it without thinking. $\endgroup$ – Kaynex Nov 10 '16 at 21:48
  • $\begingroup$ Ding ding, we have a winner. I feel dumb for not thinking to use exponent properties. Thanks @kaynex! Now to plug in some number and see how well it turns out. $\endgroup$ – chunky Nov 10 '16 at 22:19
  • $\begingroup$ Also, sorry I can't upvote. I've got the rep of a street bum. $\endgroup$ – chunky Nov 10 '16 at 22:33
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I think the first part of this is to see if you can't pull the $x$ from the exponent.

You will need to know how to split this down to only the part you are interested in: $e^{\ln(x)}$ after that you might be in a better spot to solve your problem. (though given how general it is you might still experience some difficulty!)

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    $\begingroup$ This seems more fitting to be a comment. $\endgroup$ – Clayton Nov 10 '16 at 21:41

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