Mathematics 1225A/B : 1225BW05W06_sol.pdf

• School: Western University
• Department: Mathematics
• Course: Mathematics 1225A/B
• Professor: Jennifer Mori

Document Summary

[ 1 mark ] find the value of 5log5 7. Solution: for any base b, we know that blogb a = a, so 5log5 7 = 7. [ 1 mark ] simplify log2 36 log2 4 log2 27. 27(cid:19) log2 36 log2 4 log2 27. 4 log2 27 log2 9 log2 27 log2 32 log2 33 = [ 1 mark ] if f (x) = x ln x then f (e) = Solution: we use the product rule: f (x) = 1(ln x) + x(cid:18) 1 x(cid:19) = (ln x) + 1 = 1 + ln x. Therefore f (e) = 1 + ln e = 1 + 1 = 2. [ 1 mark ] if f (x) = 32x nd f (x). Solution: we need the chain rule, along with the fact that d dx [bx] = bx ln b. We get: f (x) = 32x(ln 3)(cid:20) d dx (2x)(cid:21) = 32x(ln 3)(2) = 2(ln 3)32x.