3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '14 at 1:15
4
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You have proven that all solutions lie within the interval $(0,6)$, but how would you go about locating the solutions? $\endgroup$ – Austin Mohr Dec 30 '13 at 6:32
  • $\begingroup$ @AustinMohr, observe my assumption in the first line $\endgroup$ – lab bhattacharjee Dec 30 '13 at 6:33
  • $\begingroup$ The two sides are remarkably close on the interval [2,4]. $\endgroup$ – vadim123 Dec 30 '13 at 6:34
  • 3
    $\begingroup$ @labbhattacharjee OP does not indicate that $x$ is an integer. $\endgroup$ – Austin Mohr Dec 30 '13 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 '14 at 1:14
1
$\begingroup$

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...

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ You seem to be assuming that $x$ is an integer (or else I simply am not following the hint). $\endgroup$ – Austin Mohr Dec 30 '13 at 6:38
  • $\begingroup$ Why down vote?? $\endgroup$ – user87543 Dec 30 '13 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 '13 at 6:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.