Question 444628
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To reach a champion, *[tex \Large y\ =\ 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 64\left(\frac{1}{2}\right)^x\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1}{2}\right)^x\ =\ \frac{1}{64}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2\left(\frac{1}{2}\right)^x\ =\ \log_2\left(\frac{1}{64}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\log_2\left(\frac{1}{2}\right)\ =\ \log_2\left(\frac{1}{64}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -x\ =\ -6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 6]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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