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That was a very confusing and messy way of doing the problem. I did it in 5 lines with the same answer and did not have use a substitution
n=157^1/157. Решается двойным логарифмированием по основанию 157.
n^n^157^157=157^157 , n^157^n^157=157^157 , -> n^157=157 , ln(n)=ln157/157 , n=e^(ln157/157) , n~ 1.03273 , test , n^(n^157) , -> , (n^157)=1.03273^157 , 1.03273^157=157 , n^(n^157) -> 1.03273^(157)=157 , OK , solu. , n=~ 1.03273 ,
1
Cul🎉❤
N = 1
That was a very confusing and messy way of doing the problem. I did it in 5 lines with the same answer and did not have use a substitution
n=157^1/157. Решается двойным логарифмированием по основанию 157.
n^n^157^157=157^157 , n^157^n^157=157^157 , -> n^157=157 , ln(n)=ln157/157 , n=e^(ln157/157) , n~ 1.03273 ,
test , n^(n^157) , -> , (n^157)=1.03273^157 , 1.03273^157=157 , n^(n^157) -> 1.03273^(157)=157 , OK , solu. , n=~ 1.03273 ,
1
Cul🎉❤
N = 1