Exponential Equation From Moscow

[nycmathguy]

I found this on You Tube.

Solve for x.

3^(5)^(x) = 5^(3)^(x)

 

[Country Boy]

I first thought that notation was ambiguous. That it could be (3^5)^x= (5^3)^x or 3^(5^x)= 5^(3^x).

However it is easy to see that the first form has no solution so it must be the second.

Taking the logarithm of both sides, 5^x log(3)= 3^x log 5. Taking the logarithm again, x log(5)+ log(log(3))= x log(3)+ log(log(5)).

x(log(5)- log(3))= log(log(5))- log(log(3)).
x= (log(log(5))- log(log(3)))/(log(5)- log(3)).
x= log(log(5)/log(3))/log(5/3).

 

[nycmathguy]

I first thought that notation was ambiguous. That it could be (3^5)^x= (5^3)^x or 3^(5^x)= 5^(3^x).

However it is easy to see that the first form has no solution so it must be the second.

Taking the logarithm of both sides, 5^x log(3)= 3^x log 5. Taking the logarithm again, x log(5)+ log(log(3))= x log(3)+ log(log(5)).

x(log(5)- log(3))= log(log(5))- log(log(3)).
x= (log(log(5))- log(log(3)))/(log(5)- log(3)).
x= log(log(5)/log(3))/log(5/3).

Very good. Thanks.