If x = 2⁸ and xx=2yx^x = 2^yxx=2y, then find the value of 'y'. A111B242^424C2642^{64}264D2112^{11}211Answer: 2112^{11}211 Read Explanation: x=28x = 2^8x=28xx=2yx^x = 2^yxx=2y(28)28(2^{8})^{2^8}(28)28(am)n=a(m×n)(a^m)^n=a^{(m\times n)}(am)n=a(m×n)so28×282^{8\times 2^8}28×28= 2y2^y2yy=8×28y=8\times 2^8y=8×28y=23×28y=2^3 \times 2^8y=23×28am×an=am+na^m\times a^n=a^{m+n}am×an=am+ny=211y=2^{11}y=211 Read more in App