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

\(50^3 = 5^3 * 10^3 = 5^3 * 5^3 * 2^3\)

\(5^3\) is 125 and I know that \(2^7\) is 128.

Thus,
\(2^7 * 2^7 * 2^3 = 2^{17}\)

Answer: B

MANHATTAN GMAT OFFICIAL SOLUTION:

The underlying math facts that you need to know for this problem are the powers of 2, through 2^10. Know these powers, or be able to rederive them quickly.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
2^6 = 64
2^7 = 128
2^8 = 256
2^9 = 512
2^10 = 1,024

We are asked what integer power of 2 gives us a cube root close to 50. In other words, 50^3 should be close to that power of 2.

In equation form, we have 2^n ≈ 50^3. What is n?

One approach is to find 50^3, which equals (5^3 )(10^3 ) = 125 × 1,000 or 125,000.

Look at our list of powers of 2 up to 2^10, and let’s match up to 125,000.

125 is approximately 2^7 (= 128), while 1,000 is approximately 2^10 (= 1,024).

So 125,000 is approximately 2^7 × 2^10, or 2^17. So n = 17. No other power of 2 is even close to 125,000. By the way, 2^17 equals 131,072, but the last thing you should do is calculate that number exactly.

You could also take the original equation and multiply in 2^3:

2^n ≈ 50^3
2^3 × 2^n ≈ 2^3 × 50^3
2^(n+3) ≈ 100^3 = 1,000,000 = 1,000^2.

Since 2^10 ≈ 1,000, we know that 1,000^2 ≈ (2^10)^2 = 2^20.

So n + 3 = 20, or n = 17.

The correct answer is B.
_________________

Integer powers of 2 = 2 , 2^2, 2^3, 2^4, 2^5, 2^6, 2^7 etc.

Let, Cube root of x is closest to 50

i.e. x is closest to cube of 50 = 50*50*50 = 125000

But x must be equivalent to an Integer power of 2

2^10 = 1024
2^7 = 128
i.e. 2^17 = 128*1024 = 128000 (approx.)
i.e. 2^18 = 256*1024 = 256000 (approx.)
i.e. 2^16 = 64*1024 = 64000 (approx.)

i.e. 2^17 which is closest to 125000

i.e. Integer power of 2 must be 7

Answer: option B

Cube of 50 = 125,000
From the options:
2^16 = 2^10*2^6 = 1024*64 ≈ 64,000
2^17 = 2^10*2^7 = 1024*128 ≈ 128,000
2^16 = 2^10*2^8 = 1024*256 ≈ 256,000
Integer power of 2^17 is closest to 50^3 → Cube root of 2^17 is closest to 50.
Answer: B

Hey guys from the setup I did this fairly quickly with the LOG function. Maybe someone could check it over. I find using logs pretty easy

cubic root of 2^n = 50

2^n=50^3

Log both sides to solve for n

n*log(2) = 3*log(50)
n*log(2) = 3*[log(5)+log(10)]
0.301n=3[0.698+1]

*simplied and rounded since they are pretty close to tenths

0.3n=3*(.7+1)
0.3n=2.1+3
0.3n=5.1
n=5.1/(3/10)
n=17

Yes that is an acceptable method but only if you are comfortable with the log function and remeber the values for log 2, log 3 and log 5.

Hey guys I'm not sure if I'm misunderstanding the question but is my approach totally incorrect?

I tested the answer choices:

\((2^{17})^\frac{1}{3}=2^\frac{17}{3}=2^{5.6}=50ish\)

Given: 3√2^n=2^(n/3) ≈ 50

50 is somewhere between 2^5 (which is 32) and 2^6 (which is 64).

If 2^5 -> n/3 = 5 -> n=15
If 2^6 -> n/3 = 6 -> n=18

Now, we know that n is somewhere between 15 and 18, which are 3 units apart. 2^5 (=32) and 2^6 (=64) are 33 units apart (33=64-32+1). Thus, every 1 unit of n is 11 units (11=33/3) of 2^1k. Or 3:33 -> 1:11

Thus, 2^6 (which is 64) - 11 = 53 (closest to 50) and n (which is 18 for n^6) - 1 = 17.

Bunuel generis I chose B based on following reasoning. I solved it backwards or sidewards

Option A) 16 *3 = 48

Option B) 17*3 = 51

so here I see that 51 is closer to 50 than 48 to 50, hence B

This is how i solved it:
2^n = 50^3
2^n = (5^2 * 2^1)^3
2^n = 125*125* 2^3
2^n = 2^7 * 2^7 * 2^3
2^n = 2^(7+7+3)
2^n = 2^17
n = 17
Answer is B

A bit of a weighted average concept thrown in:

(2)^5 = 32

50

(2)^6 = 64


The integer power of 2 we are looking for as the answer will end up as a result between 5 and 6 after taking the Cube Root.

The Cube Root of an Integer is the same as raising that integer to the (1/3)rd Power

We want the resulting power, after taking the cube root, to be somewhere between (5) and (6)

(C) (2)^18

Taking the cube root we end up with: (2)^6

The answer must be A or B

(A)if we take (1/3) of the power 16, we end up with

5.3333

However, since the result we are approximating is 50 and

50 is closer to 64 (6th power)

than 32 (5th power)

we want the resulting power to be a little greater than > 5.5

(B) 17

If we take (1/3) of the power 17 we end up with:

5.666

Which will give us a value closer to the target result of 50


Answer (B)

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2^x/3 = 50
2^x = 50^3
2^x = 125 * 1000
2^x = 5^6 x 2^3
2^x = 25 x 25 x 25 2^3
2^4 = 16, 2 ^ 5 = 32 so 25 must be somewhere around 2^4.5
2^x = 2^4.5+4.5+4.5+3 = 2^16.5 =2^17

It's useful to know that

10^.3 is very close to 2.



So, 2^(x/3) = 10*10/10^.3 and

10^(.3x/3) = 10^1.7

Equating powers

.3x/3 = 1.7

.3x = 5.1

X = 5.1*(10/3) = 10*1.7

= 17

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