# Can this be solved algebraically? $2^x (6 - x) = 8x$

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.

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

• You have proven that all solutions lie within the interval $(0,6)$, but how would you go about locating the solutions? Dec 30, 2013 at 6:32
• @AustinMohr, observe my assumption in the first line Dec 30, 2013 at 6:33
• The two sides are remarkably close on the interval [2,4]. Dec 30, 2013 at 6:34
• @labbhattacharjee OP does not indicate that $x$ is an integer. Dec 30, 2013 at 6:35
• Just edited the question -- I think there might be a mistake in the problem I was looking at. Thanks for your help, everyone! Jan 3, 2014 at 1:14

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.

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

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...

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