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

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members

by VJesus12 » Sat Feb 17, 2018 7:35 am

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.

Is there an easy way to solve this PS question? Experts, can you help me? Thanks in advanced.

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Source: Beat The GMAT — Problem Solving | Sat Feb 17, 2018 7:35 am

EconomistGMATTutor: GMAT Instructor; Posts: 555; Joined: Wed Oct 04, 2017 4:18 pmFollowed by: 12 members; Thanked: 180 times

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by EconomistGMATTutor » Sat Feb 17, 2018 8:20 am

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.

Is there an easy way to solve this PS question? Experts, can you help me? Thanks in advanced.

Hello Vjesus12.

Let's take a look at your question.

We have to rewrite the power and then we just have to take a look at the exponents. The calculation is: $$8^{-x}<\frac{1}{32^3}\ \Leftrightarrow\ \ \ \frac{1}{8^x}<\frac{1}{32^3}\ \Leftrightarrow\ \ 32^3<8^x$$ $$2^{5\cdot3}<2^{3\cdot}^x\ \Leftrightarrow\ \ 2^{15}<2^{3x}\ \Leftrightarrow\ \ 15<3x\ \Leftrightarrow\ 5<x.$$ Hence, the smallest positive integer is 6.

Therefore, the correct answer is the option C.

I hope it helps you.

I'm available if you'd like a follow-up.

Regards.

GMAT Prep From The Economist
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[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members; Location: Palo Alto, CA; Thanked: 2867 times; GMAT Score:800

by [email protected] » Sat Feb 17, 2018 10:24 am

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

Contact Rich at [email protected]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8155; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Sun Jun 23, 2019 10:23 am

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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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