An algebra problem by Victor Loh

Algebra Level 3

Victor knows that $2^{3} = 8 > 2^{2} = 4$ $3^{3} = 27 > 2^{3} = 8$ $4^{3} = 64 > 2^{4} = 16$ Since $4^{3} - 2^{4} = 48 > 3^{3} - 2^{3} = 19 > 2^{3} - 2^{2} = 4,$ Victor concludes that for all integers $n \geq 2$ , $n^{3} > 2^{n}.$ However, his math teacher tells him that this only holds true up till a certain value of $n$ , that is, when $n \geq a$ where $a$ is a positive integer, $n^{3} \leq 2^{n}.$ Find the value of $a$ .

The answer is 10.

1 solution

Snehal Shekatkar
Jul 10, 2014

The primary reason that

$n^{3}<2^{n}$

for sufficiently large values of $n$ is that $n^{3}$ is a polynomial or power law function whereas $2^{n}$ is an exponential function and it is always very rapidly function of its argument. In the present case, by trial and error we can see that $10^{3}=1000<2^{10}=1024$ and hence $\boxed{a=10}$

An algebra problem by Victor Loh

The answer is 10.

1 solution

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