Lesson: Negative Exponents

Comment on Negative Exponents

Hi Brent
If x is an integer, is x|x|<2^x?

Pls explain i couldn't understand how to substitute values for |x| should i take -ve or + ve
gmat-admin's picture

In both statements you are told that x is negative. So, if you're going to plug in (test) values of x, you must use negative values.

The key thing to recognize here is that 2^x IS POSITIVE for ALL values of x (positive or negative)

For example, if x = 5, then 2^x = 2^5 = 32
And if x = -3, then 2^x = 2^(-3) = 1/8

(1) If x < 0, then x is NEGATIVE

2) Same process here as for statement 1.

Answer: D

Here's a thread discussing this question at greater length: https://gmatclub.com/forum/if-x-is-an-integer-is-x-x-2-x-144342.html

Hi Brent. Can you walk through the steps with your answer to this question:


And why/how can you flip each fraction?

Your answer to the question:

Given: 5n/(4n - x) = (0.788)^(-1)
Rewrite as: 5n/(4n - x) = 1/0.788
Flip each fraction: (4n-x)/5n = 0.788
Apply above property: 4n/5n - x/5n = 0.788
Simplify: 4/5 - x/5n = 0.788
Simplify: 0.8 - x/5n = 0.788
So: x/5n = 0.012
Multiply both sides by 5 to get: x/n = 0.06
Rewrite as fraction: x/n = 6/100 = 3/50

gmat-admin's picture

Question link: https://gmatclub.com/forum/if-5n-4n-x-0-788-then-x-n-259657.html

The main property here is as follows:
If wxyz ≠ 0 and w/x = y/z, then it must also be true that: x/w = z/y

For example, since 4/8 = 1/2, it must also be true that 8/4 = 2/1
Likewise, since 9/6 = 3/2, it must also be true that 6/9 = 2/3

So, when I got to this point: 5n/(4n - x) = 1/0.788...
...I could see that flipping both sides was going to make it much easier to solve the equation.

Does that help?


Yulia's picture

Hi Brent,

Could you please show how to solve the question by using just variables?

How did you know that x can be 2 and y can be 4?
gmat-admin's picture

Since there are no restrictions on the values of x and y, we can assign any values we wish.
I chose x = 2 and y = 4, because they're easy numbers to work with.
I could have also chosen to use x = 6.215 and y = -0.004213, and the algebraic expression would still evaluate to be -1 (but the calculations would be a lot harder to complete.)


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