Question 302789
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(x)\ =\ 7]


Since we know


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x)\ =\ y]


means that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^y\ =\ x]


We can say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2^7]


But we also know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16\ =\ 2^4]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(16x^2)\ =\ \log_2\left(2^4\left(2^7\right)^2\right)]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \log_2(2^{18})]


Use the Power Rule


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 18\log_2(2)\ =\ 18] because


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(b)\ =1]


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