Mathematics 1225A/B : 1225BW11_sol.pdf

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

Document Summary

Solution: we know that for any base b, logb ba = a, and also that ln x means loge x. Therefore ln(cid:0)e3(cid:1) = 3 and so we get [ln(e3)]2 = 32 = 9. Another approach: since ln(e3) = 3 ln e and ln e = 1, we see that [ln(e3)]2 = [3 ln e]2 = [3(1)]2 = 9. [1 mark ] simplify log4 2 + log4 8 log2 4. E: log2 4 log4 2 + log4 8 log2 4 log4(2 8) log2 4 log4 16 log2 4 log4 42 log2 22 = [1 mark ] find the slope of the tangent line to the graph of y = log3 x at the point with x = 3. Solution: the slope of the tangent line to the graph of y = f (x) at the point with x = a is given by f (a). [1 mark ] if f (x) = xln 2, nd f (x).