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I have this equation:

$3x+1 = \frac{x}{2^a}$

And I'm trying to solve for $a$ such that the expression of $a$ does not contain an $x$ in it.

No matter how much I try I can't isolate the $x$ and $a$...

Does anyone know a way to do this?

Thanks

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  • $\begingroup$ Not sure what you mean by "the expression of a does not contain an x in it", how about that ? 3x+1 = x/2^a => 2^a(3x+1)=x => 2^a = x/(3x+1) => log_2 2^a = log_2 (x/(3x+1)) => a = log_2(x/(3x+1)) $\endgroup$ – Mone Feb 18 '14 at 18:47
  • $\begingroup$ For example, if I try to solve for $a$, it will end up being a $log$ of like $x/(3x+1)$, this has $x$'s in it which I am trying to avoid... $\endgroup$ – omega Feb 18 '14 at 18:48
  • $\begingroup$ when we "solve for something" doesn't it mean to write everything else in terms of it ? (including x in this case) $\endgroup$ – Mone Feb 18 '14 at 18:50
  • $\begingroup$ This is not an abstract-algebra question, rather, an algebra-precalculus one. $\endgroup$ – Hakim Feb 18 '14 at 18:59
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$$\begin{align} 3x+1=\frac{x}{2^a} & \iff \frac1{3x+1}=\frac{2^a}x \\ & \iff 2^a=\frac{x}{3x+1} \\ & \iff \log_2(2^a)=\log_2\left[\frac{x}{3x+1}\right] \\ & \iff a=\log_2\left[\frac{x}{3x+1}\right] \,\checkmark \end{align}$$

I hope this helps.
Best wishes, $\mathcal H$akim.

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You cannot get rid of $x$, since $a$ depends on $x$. For different values of $x$ you get different values of $a$. You have

$$ a = \log_2{\frac{x}{3x+1}} $$

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