Challenger App

No.1 PSC Learning App

1M+ Downloads

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}


Related Questions:

2x² + 3y² = 6 എന്ന എലിപ്സിന്റെ എക്‌സെന്ട്രിസിറ്റി കണ്ടെത്തുക

62×104×15326×35×56= \frac {6^2 \times 10^4 \times 15^3} {2^6 \times 3^5 \times 5^6} =\rule{1cm}{0.1 pt}

x0.748=36x2.3\frac{x^{0.7}}{48} = \frac{36}{x^{2.3}}

x = ?

212+212=2n2^{12}+2^{12} =2^{n} എന്നാൽ n -ന്റെ  വില എത്ര ?

10×(23)2×(53)2=\sqrt{10\times{\sqrt{(2^3)^2}}\times\sqrt{(5^3)^2}}=