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The positive value of x that satisfies the equation = is between
A. 0 and 0.5
B. 0.5 and 1
C. 1 and 1.5
D. 1.5 and 2
E. 2 and 2.5
The two expressions in question, and , could in theory be expanded, but you’ll wind up doing a lot of algebra, only to find an equation involving , , , , and x. Solving for the positive value of x that makes the equation true is nearly impossible.
So how on Earth can you answer the question? Notice that the question does not require you to find a precise value of x; you just need a range. So plug in good benchmarks and track the value of each side of the equation.
Start with x = 1. The “equation” becomes =?= .
Compute the two sides: = 243, while = = 256. So the right side is bigger.
Now, we need another benchmark. Try x = 2. The “equation” becomes =?= .
Compute or estimate the two sides. = = 625, and multiplying in another 5 gives you something larger than 3,000 (3,125, to be precise). Meanwhile, 74 = < = 2,500, so the left side is now bigger. Somewhere between x = 1 and x = 2, then, the equation must be true.
The only benchmark left to try is 1.5, or 3/2. Plugging in, you get =?=
= = 1,024 (it’s good to know your powers of 2 this high)
For the other side, first compute the denominator: = 16. Now the numerator: = 121×11×11. 121×11 = 1,331 (this is quick to do longhand)
1,331×11 = 14,641 (also quick to do longhand)
So = 14,641/16 < 1,000. The right side is bigger.
= at what value of x?
< when x = 1
> when x = 1.5
> when x = 2
So the value at which the two sides are equal must be between 1 and 1.5.
The correct answer is C.
As x grows past 2, the larger power (5) on the left takes over, so you can see that the left side will always be bigger.
What about when x is between 0 and 1? Well, first notice that at x = 0, the two sides are again equal. If x is a tiny positive number (say, 0.001 or something), then you can ignore higher powers of x (, , etc.), and this simplifies the algebraic expansion:
For tiny positive x,
= 1 + 5(2x) + higher powers of x ˜ 1 + 10x
= 1 + 4(3x) + higher powers of x ˜ 1 + 12x
So right away, the right side is bigger than the left side. You can also check x = ½:
= = 32
= = = > 36 (= ) > 32
Again, the right side is bigger than the left side. For every x between 0 and 1, in fact, the right side is larger than the left side. This fact isn’t particularly easy to prove, but you don’t need to do so.
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