Simple GRE practice problem, but for some reason my algebraic approach is failing me, can someone point out my error?

Given: $$\theta x = x^{-3}(2x)(\frac{x}{2})(2)$$

Question: Which is greater: $\theta 8$ or $\theta 4$

I approached it by solving for theta:

$$\theta x = x^{-3}(2x)(x)$$ $$\theta x = \frac{2x^2}{x^3}$$ $$\theta x = \frac{2}{x}$$ $$\theta = \frac{2}{x^2}$$

Then plug in for $\theta 8$ and $\theta 4$: $$\theta * 8 = \frac{2}{x^2} (8) = \frac{16}{x^2}$$ $$\theta * 4 = \frac{2}{x^2} (4) = \frac{8}{x^2}$$

For any value of $x$, other than $0$, $\theta 8$ is larger. But clearly amiss here, because the opposite is true, if I plug in $\theta 8$ directly I get:

$$\theta 8 = 8^{-3}(2*8)(\frac{8}{2})(8) = \frac{1}{4}$$ $$\theta 4 = 4^{-3}(2*4)(\frac{4}{2})(4) = \frac{1}{2}$$

Now it's clear that $\theta 4$ is larger.

Ooff! for the life of me I don't see why my algebraic approach failed. How'd I get myself into this quandary? And more importantly, how do I get out using algebra?

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    $\begingroup$ $\theta(8)$ can't end up with an $x$ in it. $x=8$. $\endgroup$ – Thomas Andrews Nov 4 '14 at 22:47
  • $\begingroup$ I didn't mean to write theta(8) as a function, I meant the notation as theta*8 (a coincidentally ambiguous notation at an inopportune moment) $\endgroup$ – David Parks Nov 4 '14 at 23:30
  • $\begingroup$ But none of the above makes sense as $\theta * x$, it only makes sens as $\theta(x)$, a function. $\endgroup$ – Thomas Andrews Nov 4 '14 at 23:31
  • $\begingroup$ The practice question (out of Princeton Review) gave the equation in the form exactly as shown here, and it was a simple which is greater question with the same form shown $\theta 8$ or $\theta 4$. I was simply plugging in the equivalent form of theta (which I got from simplifying the given equation) into the statement $\theta 8$. $\endgroup$ – David Parks Nov 4 '14 at 23:37
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    $\begingroup$ But then $\theta$ would be a constant, which makes no sense, because no constant fits for all $x$. If you don't agree it means a function, why did you select Nick's answer below as "correct," when it says that $\theta$ is a function? $\endgroup$ – Thomas Andrews Nov 4 '14 at 23:39

This is a functions question, where $\theta x$ represents we are taking $x$ as an input to the function theta. Think of it as $f(x)=x^{-3}(2x)(\dfrac{x}{2})(2)$ instead. You cannot "solve" for theta because it is just notation.

  • $\begingroup$ Thanks for the answer. I see what your saying, as a function this all makes sense. I can't say I quite internalize the rational here though. I still look at it with some bewilderment. (1) I see an equation. (2) I manipulated that equation without (knowing) violating any laws of basic mathematics. (3) I now have an equivalency for theta. (4) Shouldn't I be able to replace theta with its equivalent form safely? $\endgroup$ – David Parks Nov 4 '14 at 23:34
  • $\begingroup$ I actually have an interest in math well above this level, so mucking this up really scares the bajeebers out of me, and I want to make sure I get this simple foundation straight in my head. :) $\endgroup$ – David Parks Nov 4 '14 at 23:35
  • $\begingroup$ Given how the question is written, you should think of $\theta$ as a variable as you were doing. That's why both NickC and I are guessing it's a typo and should be expressed as $\theta(x)$ rather than $\theta x$. So either you made a mistake in transcribing the question on this site, or you transcribed it correctly and are right to be confused - it was a typo wherever you saw it! $\endgroup$ – Shane Nov 5 '14 at 1:36

Maybe it should be $$\theta(x)=x^{-3}(2x)(x/2)2 = 2x^{-1} = \frac{2}{x}$$

Then obviously $\theta(4)$ > $\theta(8)$.

GRE's are not tough, but that seems too easy. Am I missing something?

  • 1
    $\begingroup$ Can I ask what a GRE is just for my own curiosity? :) $\endgroup$ – snulty Nov 5 '14 at 0:19
  • 2
    $\begingroup$ Graduate record examination. The test you take to get into grad school. $\endgroup$ – Shane Nov 5 '14 at 0:51

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