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

Guest Mar 5, 2017

edited by
Guest
Mar 5, 2017

#1**+5 **

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**

MaxWong
Mar 5, 2017

#1**+5 **

Best Answer

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**

MaxWong
Mar 5, 2017

#2**0 **

x^{2} would be 2^{32} times 2^{32 }because 2^{32} times 2^{32} equals 2^{64 }then x would be plus or minus 2^{32 }but since x is positive it would be plus 2^{32} so the question would be asking what is 2^{y }if it equals x squared 32 is 2^{5 }than you can say 2^{32 }squared is 2^{2^37 }because^{ }(2^{^2^5})^{^2^32 }is 2^{2^37 }therefore the value of y would be 2^{37 }

Guest Mar 5, 2017

#3**+5 **

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

!

asinus
Mar 5, 2017