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.
If 8^(-x) < 1/32^3, what is the smallest integer value of
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Hello Vjesus12.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.
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.
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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
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
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We are given that 8^(-x) < 1/32^3. Let's start by simplifying 8^(-x):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.
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
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