If a2+b2 = 7ab, Prove that log((a+b)/3) = 1/2(loga+logb). If x = logbc, y = logca and z = logab, Prove that xyz = x+y+z

Here are the solutions to the problems.

:
Given the equation $a^2 + b^2 = 7ab$, we need to prove that $\log \left(\frac{a+b}{3}\right) = \frac{1}{2}(\log a + \log b)$.

Step 1: Start with the given equation and add $2ab$ to both sides to form a perfect square.

$$a^2 + b^2 = 7ab$$ $$a^2 + b^2 + 2ab = 7ab + 2ab$$ $$(a+b)^2 = 9ab$$

Step 2: Take the square root of both sides.

$$\sqrt{(a+b)^2} = \sqrt{9ab}$$ $$a+b = 3\sqrt{ab}$$

(Assuming $a, b > 0$ for the logarithms to be defined and $a+b > 0$)

Step 3: Divide both sides by 3.

$$\frac{a+b}{3} = \sqrt{ab}$$

Step 4: Take the logarithm of both sides.

$$\log \left(\frac{a+b}{3}\right) = \log (\sqrt{ab})$$

Step 5: Apply the logarithm property $\log(x^n) = n \log x$.

$$\log \left(\frac{a+b}{3}\right) = \log ((ab)^{1/2})$$ $$\log \left(\frac{a+b}{3}\right) = \frac{1}{2} \log (ab)$$

Step 6: Apply the logarithm property $\log(xy) = \log x + \log y$.

$$\log \left(\frac{a+b}{3}\right) = \frac{1}{2} (\log a + \log b)$$

This completes the proof.

$$\boxed{\text{Proven: } \log \left(\frac{a+b}{3}\right) = \frac{1}{2}(\log a + \log b)}$$

:
Given $x = \log_a bc$, $y = \log_b ca$, and $z = \log_c ab$, we need to prove that $xyz = x+y+z+2$.

Step 1: Use the change of base formula $\log_p q = \frac{\log q}{\log p}$. Let's use a common base (e.g., natural logarithm or base 10 logarithm) for $\log$. Let $\log a = L_a$, $\log b = L_b$, $\log c = L_c$.

$$x = \log_a (bc) = \frac{\log(bc)}{\log a} = \frac{\log b + \log c}{\log a} = \frac{L_b + L_c}{L_a}$$ $$y = \log_b (ca) = \frac{\log(ca)}{\log b} = \frac{\log c + \log a}{\log b} = \frac{L_c + L_a}{L_b}$$ $$z = \log_c (ab) = \frac{\log(ab)}{\log c} = \frac{\log a + \log b}{\log c} = \frac{L_a + L_b}{L_c}$$

Step 2: Calculate the product $xyz$.

$$xyz = \left(\frac{L_b + L_c}{L_a}\right) \left(\frac{L_c + L_a}{L_b}\right) \left(\frac{L_a + L_b}{L_c}\right)$$ $$xyz = \frac{(L_b + L_c)(L_c + L_a)(L_a + L_b)}{L_a L_b L_c}$$

Expand the numerator:

$$(L_b + L_c)(L_c + L_a)(L_a + L_b) = (L_b L_c + L_b L_a + L_c^2 + L_c L_a)(L_a + L_b)$$ $$= L_a L_b L_c + L_a^2 L_b + L_a L_c^2 + L_a^2 L_c + L_b^2 L_c + L_a L_b^2 + L_b L_c^2 + L_a L_b L_c$$ $$= L_a^2 L_b + L_a^2 L_c + L_b^2 L_a + L_b^2 L_c + L_c^2 L_a + L_c^2 L_b + 2 L_a L_b L_c$$

So,

$$xyz = \frac{L_a^2 L_b + L_a^2 L_c + L_b^2 L_a + L_b^2 L_c + L_c^2 L_a + L_c^2 L_b + 2 L_a L_b L_c}{L_a L_b L_c}$$ $$xyz = \frac{L_a^2 L_b}{L_a L_b L_c} + \frac{L_a^2 L_c}{L_a L_b L_c} + \frac{L_b^2 L_a}{L_a L_b L_c} + \frac{L_b^2 L_c}{L_a L_b L_c} + \frac{L_c^2 L_a}{L_a L_b L_c} + \frac{L_c^2 L_b}{L_a L_b L_c} + \frac{2 L_a L_b L_c}{L_a L_b L_c}$$ $$xyz = \frac{L_a}{L_c} + \frac{L_a}{L_b} + \frac{L_b}{L_c} + \frac{L_b}{L_a} + \frac{L_c}{L_b} + \frac{L_c}{L_a} + 2$$

Step 3: Calculate the sum $x+y+z$.

$$x+y+z = \frac{L_b + L_c}{L_a} + \frac{L_c + L_a}{L_b} + \frac{L_a + L_b}{L_c}$$ $$x+y+z = \left(\frac{L_b}{L_a} + \frac{L_c}{L_a}\right) + \left(\frac{L_c}{L_b} + \frac{L_a}{L_b}\right) + \left(\frac{L_a}{L_c} + \frac{L_b}{L_c}\right)$$ $$x+y+z = \frac{L_b}{L_a} + \frac{L_c}{L_a} + \frac{L_c}{L_b} + \frac{L_a}{L_b} + \frac{L_a}{L_c} + \frac{L_b}{L_c}$$

Step 4: Compare $xyz$ and $x+y+z+2$.
From Step 2:

$$xyz = \frac{L_b}{L_a} + \frac{L_c}{L_a} + \frac{L_c}{L_b} + \frac{L_a}{L_b} + \frac{L_a}{L_c} + \frac{L_b}{L_c} + 2$$

From Step 3:

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