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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
+6765
+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
Sort:

#1
+6765
+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
#2
0

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

Guest Mar 5, 2017
#3
+6683
+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

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