if x>0, x^2=2^64 and x^x=2^y what is the value of y?
Tell me how to do it please dont just give me the answer. Prove it to 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
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 2y if it equals x squared 32 is 25 than you can say 232 squared is 22^37 because (2^2^5)^2^32 is 22^37 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.
\(\large y={2^{37}} \)
\(y=137438953472\)
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\(\large x^x=2^y\) 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
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