2 years ago

If 2^x > 4^ 15 and x is a positive integer, what is the least possible value of x?

Mathematics

1 answer:

lisov135 [29]2 years ago

5 0

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

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