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?
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.