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If x = 2⁸ and xx=2yx^x = 2^y, then find the value of 'y'.

A11

B242^4

C2642^{64}

D2112^{11}

Answer:

2112^{11}

Read Explanation:

x=28x = 2^8

xx=2yx^x = 2^y

(28)28(2^{8})^{2^8}

(am)n=a(m×n)(a^m)^n=a^{(m\times n)}

so

28×282^{8\times 2^8}= 2y2^y

y=8×28y=8\times 2^8

y=23×28y=2^3 \times 2^8

am×an=am+na^m\times a^n=a^{m+n}

y=211y=2^{11}


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