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Arrange the following in descending order
(A) $$3^{3^{3^{3}}}$$
(B) $$3^{(33)^{3}}$$
(C) $$(3^3)^{33}$$
(D) $$3^{333}$$
Choose the correct answer from the options given below :
$$(A)$$:Β $$3^{3^{3^{3}}}$$ =Β $$3^{3^{27}}$$
$$(B)$$:Β $$3^{(33)^{3}}$$ =Β $$3^{(33)^{3}}$$
$$(C)$$:Β $$(3^3)^{33}$$ =Β $$3^{99}$$
$$(D)$$:Β $$3^{333}$$
So, the powers of 3 in (A), (B), (C) and (D) are:Β $$3^{27},33^3,99,333$$ respectively.
Now, we know,Β $$3^{27}>33^3>333>99$$
So, the powers of $$3$$ in $$(A)>(B)>(D)>(C)$$
So, correct answer is option $$(A)$$
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