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.

















