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Updated on: 17 Dec 2022, 07:15
Originally posted by PriyamRathor on 17 Dec 2022, 07:05.
Last edited by Bunuel on 17 Dec 2022, 07:15, edited 1 time in total.
Last edited by Bunuel on 17 Dec 2022, 07:15, edited 1 time in total.
Formatted.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
72% (00:51) correct
28%
(00:43)
wrong
based on 58
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Simplify the given expression \(\frac{3 ^{(3-1)^3}}{ 3^{3^{(3-1)}}}=\)
A. 1
B. 1/3
C. 1/9
D. 1/27
E. 1/81
Quant.jpg [ 25.06 KiB | Viewed 1565 times ]
gmatphobia chetan2u
Also please explain the rule behind this
I solved it this way :-
9^3 /27^2 == 9 x 9 x 9 / 27 x 27 = 1
But the answer is 1/3
Please guide
A. 1
B. 1/3
C. 1/9
D. 1/27
E. 1/81
Attachment:
Quant.jpg [ 25.06 KiB | Viewed 1565 times ]
Also please explain the rule behind this
I solved it this way :-
9^3 /27^2 == 9 x 9 x 9 / 27 x 27 = 1
But the answer is 1/3
Please guide
17 Dec 2022, 07:19
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PriyamRathor
\(\frac{3 ^{(3-1)^3}}{ 3^{3^{(3-1)}}}=\frac{3 ^{2^3}}{ 3^{3^2}}=\frac{3^8}{3^9}=\frac{1}{3}\).
Answer: B.
\(a^{m^n} = a^{(m^n)}\) and not \((a^m)^n\) (if exponentiation is indicated by stacked symbols, the rule is to work from the top down).
On the other hand, \((a^m)^n=a^{mn}\).
8. Exponents and Roots of Numbers
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17 Dec 2022, 07:37
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PriyamRathor
Hi
\(\frac{3 ^{(3-1)^3}}{ 3^{3^{(3-1)}}}=\) \(\frac{3 ^{(2)^3}}{ 3^{3^{(2)}}}=\)
Tackle the power first: \(2^3=8…….3^2=9\)
Bunuel has already given the rule. \(a^m^n=a^{(m^n)}\)
So \(\frac{3 ^{8}}{ 3^{9}}=\frac{1}{3}\)
