Germany | Math Olympiad Exponential Equation | How to solve!!

Germany | Math Olympiad Exponential Equation | How to solve!!
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A very nice Square Root Simplification | Math Olympiad | How to solve!!
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[@thelifeclass]

Interesting 🎉

[@payoo_2674]

3 real solutions
Use the Lambert W function W(■*e^■) = ■
4^x = (2*x)^32
ln(2^(2*x)) = ln((2*x)^32)
2*x*ln(2) = 32*ln|2*x| ===> two cases
1st case: x > 0
2*x*ln(2) = 32*ln(2*x)
ln(2*x)*(2*x)^(-1) = ln(2)/32
ln(2*x)*e^ln((2*x)^(-1)) = ln(2)/32
ln(2*x)*e^(-ln(2*x)) = ln(2)/32
-ln(2*x)*e^(-ln(2*x)) = -ln(2)/32
W(-ln(2*x)*e^(-ln(2*x))) = W(-ln(2)/32)
-ln(2*x) = W(-ln(2)/32)
ln(2*x) = -W(-ln(2)/32)
2*x = e^(-W(-ln(2)/32))
x = 1/2*e^(-W(-ln(2)/32)) ===> -1/e < -ln(2)/32 < 0 ===> 2 real solutions
x1 = 1/2*e^(-W_[0](-ln(2)/32)) = 0.5111964701028901603263758399247002841884182559566432258864139488...
in WolframAlpha: 1/2*e^(-productlog(0,-ln(2)/32))
x2 = 1/2*e^(-W_[-1](-ln(2)/32)) = 128
in WolframAlpha: 1/2*e^(-productlog(-1,-ln(2)/32))
2nd case: x < 0
2*x*ln(2) = 32*ln(-2*x)
ln(-2*x)*(2*x)^(-1) = ln(2)/32
-ln(-2*x)*(2*x)^(-1) = -ln(2)/32
ln(-2*x)*(-2*x)^(-1) = -ln(2)/32
ln(-2*x)*e^ln((-2*x)^(-1)) = -ln(2)/32
ln(-2*x)*e^(-ln(-2*x)) = -ln(2)/32
-ln(-2*x)*e^(-ln(-2*x)) = ln(2)/32
W(-ln(-2*x)*e^(-ln(-2*x))) = W(ln(2)/32)
-ln(-2*x) = W(ln(2)/32)
ln(-2*x) = -W(ln(2)/32)
-2*x = e^(-W(ln(2)/32))
x = -1/2*e^(-W(ln(2)/32)) ===> ln(2)/32 > 0 ===> 1 real solution
x3 = -1/2*e^(-W_[0](ln(2)/32)) = -0.489508467478892306161291275005825034374045024443005838132688541...
in WolframAlpha: -1/2*e^(-productlog(0,ln(2)/32))

[@MasterTMathsClass]

Nice 🎉