Question 226059
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^r\ +\ 2^r\ +\ 2^r\ +\ 2^r\ +\ =\ 26]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(2^r)\ =\ 26]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^r\ =\ 6.5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(2^r)\ =\ \log_2(6.5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\cdot\log_2(2)\ =\ \log_2(6.5)]


But *[tex \Large \log_b(b)\ =\ 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \log_2(6.5)]


which is an exact answer.  If you have a capable enough calculator or use Excel you can get the numeric approximation *[tex \Large r\ \approx\ 2.70044]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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