EECS 1019 Study Guide - Midterm Guide: Arithmetic Progression, Opata Language
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EECS 1019 Full Course Notes
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[2+2 points] functions: find g 1(3) given g(x) = 3x+1. Solution: we have to nd the value of x such that g(x) = 3. So we can substitute g(x) = 3 and solve for x. 3x = 8 x = 8/3: let f be the function f (x) = ax2 . 2)2 a(2a : = 2a . [2+2 points] logarithms: let x = 2logb 3 and y = 3logb 2. = logb 3 logb 2 and logb y = logb 3logb 2. So logb x = logb y and therefore x = y and thus x y = 0: if logb a logc a = 19. 99 and b c = ck, compute k. Solution: let us change the logarithm base to c in the numerator. Note that logb a = logc a logc b . So logb a logc a logc a logc a logc b.