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09 Jul 2026, 00:04
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We have this inequality: 3000 > 1/3n > 30, and n must be an integer. The tricky part is that 1/3n makes people nervous due to negative exponents, but here’s the trick: 1/3n is really 3^(-n). So we are asking which powers of 3 fall strictly between 30 and 3000.
That makes things easier. Let’s list the powers of 3 and see where they fit:
31 = 3; too small
32 = 9; still too small
33 = 27; close but not over 30
34 = 81; now we're in range.
35 = 243; good
36 = 729; good
37 = 2187; still under 3000, good
38 = 6561; too high over 3000
The powers that fit between 30 and 3000 are 34, 35, 36, and 37. Remember that 1/3n = 3^(-n). So if 3^(-n) equals one of those four powers, then -n = 4, 5, 6, or 7. Flipping the sign gives us n = -4, -5, -6, or -7.
Our candidates are -7, -6, -5, and -4. The last step is to find the median. Since we have four numbers, we average the two in the middle. Lined up, that’s -6 and -5. The average is:
(-6 + -5)/2 = -11/2
Thus, the median is -11/2, which is answer D. The whole problem relies on correctly flipping that negative exponent at the beginning. Once we convert 1/3n to 3^(-n) the rest is just arithmetic.
That makes things easier. Let’s list the powers of 3 and see where they fit:
31 = 3; too small
32 = 9; still too small
33 = 27; close but not over 30
34 = 81; now we're in range.
35 = 243; good
36 = 729; good
37 = 2187; still under 3000, good
38 = 6561; too high over 3000
The powers that fit between 30 and 3000 are 34, 35, 36, and 37. Remember that 1/3n = 3^(-n). So if 3^(-n) equals one of those four powers, then -n = 4, 5, 6, or 7. Flipping the sign gives us n = -4, -5, -6, or -7.
Our candidates are -7, -6, -5, and -4. The last step is to find the median. Since we have four numbers, we average the two in the middle. Lined up, that’s -6 and -5. The average is:
(-6 + -5)/2 = -11/2
Thus, the median is -11/2, which is answer D. The whole problem relies on correctly flipping that negative exponent at the beginning. Once we convert 1/3n to 3^(-n) the rest is just arithmetic.
09 Jul 2026, 00:28
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30 < 1/ 3^n < 3000
30 < 3^-(n) < 3000
we know 3^3 = 27
so n must be greater then 3. means starting value of n must be 4.
now we need to find the upper limit of 3 that is less than 3000.
3^7 = 2187 and 3^8 will be 6561. so n cant be 8.
now n= 4,5,6,7
but n must be in negative sign. because our range is in positive numbers. and we have 3 raise to negative value. so we need to make sure the n should be negative then those two negatives will give us positive value that falls within our range.
so n = -4,-5,-6,-7
we need median= -5-6/2= -11/2
choice D
30 < 3^-(n) < 3000
we know 3^3 = 27
so n must be greater then 3. means starting value of n must be 4.
now we need to find the upper limit of 3 that is less than 3000.
3^7 = 2187 and 3^8 will be 6561. so n cant be 8.
now n= 4,5,6,7
but n must be in negative sign. because our range is in positive numbers. and we have 3 raise to negative value. so we need to make sure the n should be negative then those two negatives will give us positive value that falls within our range.
so n = -4,-5,-6,-7
we need median= -5-6/2= -11/2
choice D
Bunuel
09 Jul 2026, 00:43
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Since 3000>1/3^n >30
But 1/3^n is also 3^-n
Therefore 30 < 3^-n < 3000
3^3 & 3^4 are 27 and 81
Therefore since3^-n >30
in 3^-n, -n = or > 4
3^7 & 3^8, are 2187 & 6561
Therefor -n < 7 or =
Possible values would be -7, -6, -5 , -4
Median is - 11/2
Hence d
But 1/3^n is also 3^-n
Therefore 30 < 3^-n < 3000
3^3 & 3^4 are 27 and 81
Therefore since3^-n >30
in 3^-n, -n = or > 4
3^7 & 3^8, are 2187 & 6561
Therefor -n < 7 or =
Possible values would be -7, -6, -5 , -4
Median is - 11/2
Hence d
Bunuel
