To show that (1)/(x) xⁿ = (x xⁿ)/(_x (x-1)).

Here's the solution for 1(b)i and 1(b)ii:

1(b)i)
The question asks to show that $\log_{\frac{1}{x}} x^n = \frac{\log_x x^n}{\log_x (x-1)}$.
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 $\log_b a = \frac{\log_c a}{\log_c b}$. Applying this to the Left Hand Side (LHS) with $b = \frac{1}{x}$, $a = x^n$, and choosing $c=x$, we get $\log_{\frac{1}{x}} x^n = \frac{\log_x x^n}{\log_x (\frac{1}{x})}$.
We will proceed by showing this standard identity, assuming the question intended the denominator to be $\log_x (\frac{1}{x})$ instead of $\log_x (x-1)$, 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.

$$\text{LHS} = \log_{\frac{1}{x}} x^n$$

Step 2: Apply the change of base formula $\log_b a = \frac{\log_c a}{\log_c b}$ using base $c=x$.

$$\text{LHS} = \frac{\log_x x^n}{\log_x (\frac{1}{x})}$$

Step 3: Simplify the numerator and denominator using the logarithm properties $\log_k k^p = p$ and $\log_k (\frac{1}{k}) = \log_k (k^{-1}) = -1$.

$$\log_x x^n = n$$ $$\log_x (\frac{1}{x}) = -1$$

Step 4: Substitute these simplified terms back into the LHS expression.

$$\text{LHS} = \frac{n}{-1} = -n$$

Step 5: Now, consider the Right Hand Side (RHS) of the intended identity.

$$\text{RHS} = \frac{\log_x x^n}{\log_x (\frac{1}{x})}$$

Step 6: Simplify the numerator and denominator as in Step 3.

$$\log_x x^n = n$$ $$\log_x (\frac{1}{x}) = -1$$

Step 7: Substitute these simplified terms back into the RHS expression.

$$\text{RHS} = \frac{n}{-1} = -n$$

Since $\text{LHS} = -n$ and $\text{RHS} = -n$, we have shown that $\text{LHS} = \text{RHS}$.
Therefore, $\boxed{\log_{\frac{1}{x}} x^n = \frac{\log_x x^n}{\log_x (\frac{1}{x})}}$.

1(b)ii)
The question asks to show that $\log_a (a+b)^2 = 2 + \log_a (1 + \frac{2b}{a} + \frac{b^2}{a^2})$.

Step 1: Start with the Right Hand Side (RHS) of the equation.

$$\text{RHS} = 2 + \log_a \left(1 + \frac{2b}{a} + \frac{b^2}{a^2}\right)$$

Step 2: Recognize that the expression inside the logarithm is a perfect square: $1 + \frac{2b}{a} + \frac{b^2}{a^2} = \left(1 + \frac{b}{a}\right)^2$.

$$\text{RHS} = 2 + \log_a \left(\left(1 + \frac{b}{a}\right)^2\right)$$

Step 3: Rewrite the term inside the parenthesis with a common denominator.

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