to show that (1)/(x) x^n = (x x^n)/(_x (x-1)).

Here's the solution for 1(b)i and 1(b)ii: 1(b)i) The question asks to show that _(1)/(x) x^n = (_x x^n)/(_x (x-1)). It appears there might be a typo in the denominator of the Right Hand Side (RHS). A standard logarithmic identity (the change of base formula) states _b a = (_c a)/(_c b). Applying this to the Left Hand Side (LHS) with b = (1)/(x), a = x^n, and choosing c=x, we get _(1)/(x) x^n = (_x x^n)/(_x (1)x). We will proceed by showing this standard identity, assuming the question intended the denominator to be _x ((1)/(x)) instead of _x (x-1), as the latter would only hold for specific values of x rather than as a general identity. Step 1: Start with the Left Hand Side (LHS) of the intended identity. LHS = _(1)/(x) x^n Step 2: Apply the change of base formula _b a = (_c a)/(_c b) using base c=x. LHS = (_x x^n)/(_x (1)x) Step 3: Simplify the numerator and denominator using the logarithm properties _k k^p = p and _k ((1)/(k)) = _k (k^-1) = -1. _x x^n = n _x ((1)/(x)) = -1 Step 4: Substitute these simplified terms back into the LHS expression. LHS = (n)/(-1) = -n Step 5: Now, consider the Right Hand Side (RHS) of the intended identity. RHS = (_x x^n)/(_x (1)x) Step 6: Simplify the numerator and denominator as in Step 3. _x x^n = n _x ((1)/(x)) = -1 Step 7: Substitute these simplified terms back into the RHS expression. RHS = (n)/(-1) = -n Since LHS = -n and RHS = -n, we have shown that LHS = RHS. Therefore, _(1)/(x) x^n = (_x x^n)/(_x (1)x). 1(b)ii) The question asks to show that _a (a+b)^2 = 2 + _a (1 + (2b)/(a) + (b^2)/(a^2)). Step 1: Start with the Right Hand Side (RHS) of the equation. RHS = 2 + _a (1 + (2b)/(a) + (b^2)/(a^2)) Step 2: Recognize that the expression inside the logarithm is a perfect square: 1 + (2b)/(a) + (b^2)/(a^2) = (1 + (b)/(a))^2. RHS = 2 + _a ((1 + (b)/(a))^2) Step 3: Rewrite the term inside the parenthesis with a common denominator. RHS = 2