2^(3^x) = 3^(2^x)

Solve for x over the real numbers:
2^(3^x) = 3^(2^x)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
log(2) 3^x = log(3) 2^x

Take the natural logarithm of both sides and use the identities log(a b) = log(a)+log(b) and log(a^b) = b log(a):
log(3) x+log(log(2)) = log(2) x+log(log(3))

Subtract x log(2)+log(log(2)) from both sides:
(log(3)-log(2)) x = log(log(3))-log(log(2))

Divide both sides by log(3)-log(2):
Answer: |  x = (log(log(3))-log(log(2)))/(log(3)-log(2))=1.135882567........

Guten Morgen ,

hier noch ein Lösungsweg !

Gruß radix !

Hallo, ich gebe auch meinen Senf dazu:

2^(3^x) = 3^(2^x)

2^(3^x) = 3^(2^x)

3^x * lg2 = 2^x * lg3

3^x / 2^x = lg 3 / lg2

(3/2)^x = lg3 / lg2

 x * lg(1,5) = lg3 / lg2

x = lg3 / (lg2 * lg1,5)

x = 9.00080168214

Bin mir aber nicht ganz sicher. asinus :- )

Danke radix, jetzt stimmts auch bei mir!

2^(3^x) = 3^(2^x)

2^(3^x) = 3^(2^x)

3^x * lg2 = 2^x * lg3

3^x / 2^x = lg 3 / lg2

(3/2)^x = lg3 / lg2

x * lg 1,5 = lg (lg3 / lg2)

x = lg (lg3 / lg2) / lg 1,5

x = 1,13588256791

2^(3^1,13588256791) = 3^(2^1,13588256791)

             11,181190789 = 11,181190789

Gruß asinus :- )

  !