Grade 9 Math help needed please

Use logarithms.

x^2 = 2^64 | Take natural log both sides

2 ln x = 64 ln 2 | /2

ln x = 64 ln 2 / 2 | e^ both sides

x = e^(64ln 2/2) = 2^(64/2) = 2^32

x^x = 2^y | natural log again......

x ln x = y ln 2 | substitution

2^32 ln 2^32 = y ln 2 | log property

32 * 2^32 ln 2 = y ln 2 | multiply

2^37 ln 2 = y ln 2 | / ln 2

y = 2^37

x² would be 2³² times 2³² because 2³² times 2³² equals 2⁶⁴  then x would be plus or minus 2³² but since x is positive it would be plus 2³² so the question would be asking what is 2^y if it equals x squared 32 is 2⁵ than you can say 2³² squared is 2^{2^37} because (2^{^2^5})^{^2^32} is 2^{2^37} therefore the value of y would be 2³⁷ 

if x>0, x^2=2^64 and x^x=2^y what is the value of y? 

$x^2=2^{64}$              square root

$\sqrt{x^2}=\sqrt{2^{64}}$       calculate

$x=2^{32}$               result for x

$\large x^x=2^y$            output equation.  x-value

$\large x^x=(2^{32})^{2^{32}}=2^{32\times 2^{32}}=2^{2^5\times 2^{32}}=2^{2^{37}}=2^y$

$\large2^{2^{37}}=2^y$          output equation. x is used

If the basis of two powers is equal, the exponents are equal.

$\large y={2^{37}}$

$y=137438953472$

  !

$\large x^x=2^y$    output equation. x and y used

$\Large(2^{32})^{(2^{32})}=2^{137438953472}$    x and y used. calculate

$\large32\times2^{32}=137438953472$  q.e.d

  !