Solve the system of equations: xy = 64 and x y + y x = (5)/(2)

Step 1: Solve 11(i) for $x$ and $y$.
We are given two equations:

1. $xy = 64$
2. $\log_x y + \log_y x = \frac{5}{2}$

For equation (2), use the change of base formula for logarithms, $\log_y x = \frac{1}{\log_x y}$.
Let $k = \log_x y$. Then the equation becomes:

$$k + \frac{1}{k} = \frac{5}{2}$$

Multiply by $2k$ to clear the denominators:

$$2k^2 + 2 = 5k$$

Rearrange into a quadratic equation:

$$2k^2 - 5k + 2 = 0$$

Factor the quadratic equation:

$$(2k-1)(k-2) = 0$$

This gives two possible values for $k$:

$$2k-1 = 0 \implies k = \frac{1}{2}$$ $$k-2 = 0 \implies k = 2$$

Case 1: $k = \frac{1}{2}$
Since $k = \log_x y$, we have $\log_x y = \frac{1}{2}$.
Convert to exponential form: $y = x^{1/2} = \sqrt{x}$.
Substitute this into the first equation $xy = 64$:

$$x(\sqrt{x}) = 64$$ $$x^{1}x^{1/2} = 64$$ $$x^{3/2} = 64$$

Raise both sides to the power of $\frac{2}{3}$:

$$x = 64^{2/3}$$ $$x = (4^3)^{2/3}$$ $$x = 4^2$$ $$x = 16$$

Now find $y$:

$$y = \sqrt{x} = \sqrt{16} = 4$$

So, one solution is $(x,y) = (16,4)$.

Case 2: $k = 2$
Since $k = \log_x y$, we have $\log_x y = 2$.
Convert to exponential form: $y = x^2$.
Substitute this into the first equation $xy = 64$:

$$x(x^2) = 64$$ $$x^3 = 64$$

Take the cube root of both sides:

$$x = \sqrt[3]{64}$$ $$x = 4$$

Now find $y$:

$$y = x^2 = 4^2 = 16$$

So, another solution is $(x,y) = (4,16)$.

The values of $x$ and $y$ are $\boxed{x=16, y=4 \text{ or } x=4, y=16}$.

Step 2: Prove 11(ii) that if $a^x = b^y = (ab)^{xy}$, then $x+y=1$.
Let $a^x = b^y = (ab)^{xy} = K$.
From $a^x = K$, we have $a = K^{1/x}$.
From $b^y = K$, we have $b = K^{1/y}$.
Substitute these expressions for $a$ and $b$ into $(ab)^{xy} = K$:

$$(K^{1/x} \cdot K^{1/y})^{xy} = K$$

Using the exponent rule $K^m \cdot K^n = K^{m+n}$:

$$(K^{\frac{1}{x} + \frac{1}{y}})^{xy} = K$$

Combine the exponents in the parenthesis:

$$(K^{\frac{y+x}{xy}})^{xy} = K$$

Using the exponent rule $(K^m)^n = K^{mn}$:

$$K^{\frac{y+x}{xy} \cdot xy} = K$$ $$K^{y+x} = K^1$$

Assuming $K \neq 1$ (which implies $a \neq 1$ and $b \neq 1$), we can equate the exponents:

$$y+x = 1$$

Therefore, $x+y=1$ is proven.

Step 3: Prove 11(iii) that $\log AB = \log A + \log B$ and $\log \frac{A}{B} = \log A - \log B$.

Proof for $\log AB = \log A + \log B$:
Let $\log_b A = x$ and $\log_b B = y$.
By the definition of logarithms, this means $A = b^x$ and $B = b^y$.
Multiply $A$ and $B$:

$$AB = b^x \cdot b^y$$

Using the exponent rule $b^m \cdot b^n = b^{m+n}$:

$$AB = b^{x+y}$$

Convert this exponential form back to logarithmic form:

$$\log_b (AB) = x+y$$

Substitute back the original definitions of $x$ and $y$:

$$\log_b (AB) = \log_b A + \log_b B$$

When the base $b$ is not specified, it is assumed to be a common base (e.g., 10 or $e$), so $\log AB = \log A + \log B$.

Proof for $\log \frac{A}{B} = \log A - \log B$:
Let $\log_b A = x$ and $\log_b B = y$.
By the definition of logarithms, this means $A = b^x$ and $B = b^y$.
Divide $A$ by $B$:

$$\frac{A}{B} = \frac{b^x}{b^y}$$

Using the exponent rule $\frac{b^m}{b^n} = b^{m-n}$:

$$\frac{A}{B} = b^{x-y}$$

Convert this exponential form back to logarithmic form:

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