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if 2^x > 4^ 15 and x is a positive integer, what is the least possible value of x?
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if 2^x > 4^ 15 and x is a positive integer, what is the least possible value of x?
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Mathematics
7 months
2021-07-20T20:35:36+00:00
2021-07-20T20:35:36+00:00 1 Answers
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Answer:
x = 31
Step-by-step explanation:
2^x > 4^ 15
Rewrite 4 as 2^2
2^x > 2^2^ 15
We know a^b^c = a^(b*c)
2^x > 2^ (2*15)
2^x > 2^30
Since the bases are the same, the exponents must follow the inequality
x> 30