A Radical Exponential Equation | Bonus at the End

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Комментарии
Автор

with sqrt(n^x)=n (or with the whole cascade again) also x=2. And if we take the k-th root each time, then x=k.

mystychief
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Sometimes it's good to take a pause.... Thanks for the video!

manwork
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square both sides first, then Log2 ( base 2 on both sides) then got (2^x)^1/2=4, square both sides and Log 2 on both sides again, then done x=2

ecaiusa
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sqrt[2^f(x)] = 2 means that f(x) = sqrt(2^x) = 2. Thus 2^x = 4, so xln2 = ln4 = ln2^2 = 2ln2. Therefore x = 2ln2/ln2 = 2.

toveirenestrand
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Complex solutions should be x=4+4*k*pi*i/ln2, k an integer

vaggelissmyrniotis