An ordinary inequality.

Algebra Level 3

How many positive integers n n satisfy the following inequality?

2 n 64 × 2 6 n 1 \Large 2^{n} \leq 64 \times 2^{\frac{6}{n-1}}


The answer is 6.

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1 solution

Kelvin Hong
Aug 25, 2017

2 n 2 6 n n 1 2^n \leq 2^{\frac{6n}{n-1}}

n 6 n n 1 n \leq \frac{6n}{n-1}

Because n > 0 n>0

1 6 n 1 1 \leq \frac{6}{n-1}

Also n 1 > 0 n-1 >0

n 1 6 n-1 \leq 6

n 7 n \leq 7 To make sense of the denominator , n 1 n \neq 1

n = 2 , 3 , 4 , 5 , 6 , 7 n= 2,3,4,5,6,7

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