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I've been working on this problem for a few days, but I haven't been able to find $x$ algebraically. (Maybe I'm missing something obvious?)

$2^x (6 - x) = 8x$

Using a MATLAB program, I found the solutions $ x = 2, 3, 4 $.

I tried using Wolfram|Alpha, but since it didn't provide steps, I assume it solved the equation numerically too. Which is not very elegant.

Follow up: I guess the problem I was looking at was not well written, and should have asked for integer solution.

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3 Answers 3

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Assuming $x$ to be a positive integer we have, $$2^{x-3}=\frac x{6-x}$$

For real $x,$ the Left hand Side is strictly positive

If $x>0,$ we need $\displaystyle 6-x>0\iff x<6\iff 0<x<6,$ observe that $x=2,3,4$ are actually solutions

If $x<0,$ we need $\displaystyle 6-x<0\iff x>6$ which is impossible

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  • $\begingroup$ You have proven that all solutions lie within the interval $(0,6)$, but how would you go about locating the solutions? $\endgroup$ Dec 30, 2013 at 6:32
  • $\begingroup$ @AustinMohr, observe my assumption in the first line $\endgroup$ Dec 30, 2013 at 6:33
  • $\begingroup$ The two sides are remarkably close on the interval [2,4]. $\endgroup$
    – vadim123
    Dec 30, 2013 at 6:34
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    $\begingroup$ @labbhattacharjee OP does not indicate that $x$ is an integer. $\endgroup$ Dec 30, 2013 at 6:35
  • $\begingroup$ Just edited the question -- I think there might be a mistake in the problem I was looking at. Thanks for your help, everyone! $\endgroup$
    – Kevin Chen
    Jan 3, 2014 at 1:14
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Hard to tell of $x$ should be an integer or not. However, if this is specified then, I would partition the right hand side as $$ 2^x = \frac{48}{6-x} - 8 $$ So $6-x$ must divide 48 and $6-x>0$. So only possible values are $x=2,3,4,5$. Saves one choice of $x$.

If nothing is known about $x$ then not much can be said.

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  • $\begingroup$ I think you are right; $x$ should probably be an integer. I've edited the question to reflect this. Thanks for your help! $\endgroup$
    – Kevin Chen
    Jan 3, 2014 at 1:15
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Partial answer :

Hints :

$2^x(6-x)=8x$ can be seen as :

$$2^x(6-x)=2.(4x)$$

$$2^x(6-x)=4.(2x)$$

$$2^x(6-x)=8(x)$$

You have to check which could be the possible case...

Only two of these three equations makes sense...

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    $\begingroup$ You seem to be assuming that $x$ is an integer (or else I simply am not following the hint). $\endgroup$ Dec 30, 2013 at 6:38
  • $\begingroup$ Why down vote?? $\endgroup$
    – user87543
    Dec 30, 2013 at 6:39
  • $\begingroup$ @AustinMohr : I am not assuming $x$ is an integer... I am just saying one integer solution can be drawn from this idea.. $\endgroup$
    – user87543
    Dec 30, 2013 at 6:50

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