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not really sure I'm doing this right but anywho...

using simpler numbers if R = 2^2
Then R^R = 2^8

Difference of 6 on the exponents

So 81 + 6 = 87

E

curious if that makes any sense.
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Don't get freaked by the number of exponents.

((x)^(y))^(z) = (x)^(yz) property

so (3^81)^(3^81) = 3^(81*3^81)

and 3^4 =81

x^y*x^z = x^(y+z)

so 3^(3^4*3^81) = 3^(3^85)

thus s = 3^85

If someone could tell me how to do superscripts I could possibly make this easier to read.
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This problem can be solved very quickly using logarithm (although it is not in the syllabus of GMAT)
R=3^81
log (R) = 81 log 3
And since,
R^R = 3^s
therefore, R log R = s log 3
Now substituting the value of log R from eqn. 1,
R X (81 log 3) = s log 3
Hence, s = 3^81 X 3^4
= 3^ 85
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One more way - Exponential -
Given - R = 3^81.....(a)
R^R = 3^S
R = 3 ^(S/R)............(b)
(a) = (b)
we get S/R = 81 i.e. S=81R i.e. S=81 * 3^81
hence S = 3^4 + 3^81 = 3^85
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Given that R = 3^81

Powering both sides by R

R^R = 3^81R............... (1)

Already given that R^R = 3^S......... (2)

Equating (1) & (2)

S = 81R
= 3^4 . 3^81
= 3^85 = Answer = D
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here's another way....

R^R = (3^81)^(3^81) -->
(27^27)^(27^27)
(27)^(27^28)
(3^3)^[3^(28*3)]
3^[3^(1+28*3)]
3^(3^85)
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R=3^81 -----given

R^R=3^S ---given

=> (3^81)^R = 3^S
=> 3^(81R) = 3^S
=> 81R = 3^S
=> 3^4 x 3^81 = 3^S
=> s=52
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R=3^81

R^R=3^S

(3^81)^R = 3^S

So, 3^(81R) = 3^S

81R = S

=> S=(3^4)(3^81)

S=3^85

(D)
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What is this concept called exactly and are there any videos on youtube explaining the principles?

Kind regards.
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