What is the log base 2 of the enclosed expression? (Variant 2)

Algebra Level pending

This problem’s question: {\color{#D61F06}\text{This problem's question:}} What is the log base 2 of 1 − 262537412640768744 e − π 163 − 196884 e − 2 π 163 + 103378831900730205293632 e − 3 π 163 1-262537412640768744 e^{-\pi \sqrt{163}}-196884 e^{-2 \pi \sqrt{163}}+103378831900730205293632 e^{-3 \pi \sqrt{163}} ?

The usual Brilliant real number allowance is applicable.

This is not a home-work problem. It can not be done on the average scientific calculator. To do it in floating point would require IEEE octuple-precision floating point. It is not a trivial 1+1 problem.


The answer is -195.30340397407311578.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

This is another difficult calculation problem. The answer is about -195.303403974073115789334667088444455502287.

This one causes Wolfram/Alpha interpretation of expression problems 

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bc -l
scale=100
p=4*a(1)
z=sqrt(163)
l(1-262537412640768744*e(-p*z)-196884*e(-2*p*z)+103378831900730205293632*e(-3*p*z))/l(2)
-195.303403974073115790137300641527523776325017791968637950070504635\
3778557140641272791609396650920011101
^D

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