The cube root of what integer power of 2 is closest to 50?

AAPL wrote: ↑

Sun Jan 19, 2020 8:12 am

Manhattan Prep

The cube root of what integer power of 2 closest to 50?

A. 16
B. 17
C. 18
D. 19
E. 20

OA B

Say the integer power of 2 closest to 50 is n; thus, we have \(\sqrt[3]{^{2^n}}= 50\)

\({2^{n/3} = 50}\)

We know that 2^5 = 32 < 50 and 2^6 = 64 > 50

Thus, 5 < n/3 < 6 => 15 < n < 18. Thus, n could be 16 or 17. Still we have two options A and B.

• Taking n = 16:

\({2^{n/3} = 50 => 2^{16/3} = 2^5*\sqrt[3]{2} = 32*\sqrt[3]{2}}\)

• Taking n = 17:

\({2^{n/3} = 50 => 2^{17/3} = (2^6)/(\sqrt[3]{2}) = 64/\sqrt[3]{2}}\)

Now we have to compare \(32*\sqrt[3]{2}\) and \(64/\sqrt[3]{2}\). Since 32 is 18 away from 50 and 64 is 14 away from 50, the value of \(64/\sqrt[3]{2}\) must be closer to 50 than \(32*\sqrt[3]{2}\) is closer to 50. Thus, n = 17.

The correct answer: B

Hope this helps!

-Jay
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AAPL wrote: ↑

Sun Jan 19, 2020 8:12 am

Manhattan Prep

The cube root of what integer power of 2 closest to 50?

A. 16
B. 17
C. 18
D. 19
E. 20

OA B

We can translate the problem into an equation where x is the integer sought:

^3√(2^x) = 50

If we raise both sides to the third power, we have:

2^x = 50^3

2^x = 125,000

Recall that 2^10 = 1,024 ≈ 1,000. So:

2^16 = 2^6 x 2^10 ≈ 64 x 1,000 = 64,000

2^17 = 2^7 x 2^10 ≈ 128 x 1,000 = 128,000

We see that 2^17 is closest to 125,000, so x = 17.

Answer: B