Hi VJesus12,

We're told that If 8^(-X) < 1/(32^3). We're asked for the SMALLEST integer value of X that will fit this inequality. Since both 8 and 32 are 'powers of 2', we can rewrite this equation with 2 as the 'base number'

8 = 2^3
32 = 2^5

(2^3)^(-X) < 1/[2^5]^3]

Next, since we're raising a 'power to a power', we multiply the exponents:

2^(-3X) < 2^(-15)

Since the bases are now the same, we can ignore them and focus on the exponents and inequality:

-3X < -15
3X > 15
X > 5

The smallest integer greater than 5 is 6.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

VJesus12 wrote:If 8^(-x) < 1/32^3, what is the smallest integer value of x?

A. 4
B. 5
C. 6
D. 8
E. 9

The OA is C.

We are given that 8^(-x) < 1/32^3. Let's start by simplifying 8^(-x):

8^(-x) = (2^3)^(-x) = 2^(-3x)

We can also simplify 1/32^3:

1/32^3 = 1/(2^5)^3 = 1/(2^15) = 2^(-15)

Thus, 2^(-3x) < 2^(-15).

Since our bases are equal and are greater than 1, we can drop the bases and solve the inequality in terms of the exponents.

-3x < -15

x > 5 (Note: we switch the inequality sign because we divide both sides by -3, a negative number.)

Since x is greater than 5, the smallest integer value of x is 6.

Answer: C