Proving logarithm question

Question

Prove: $$\log_a (bc)\times \log_b (ac)\times \log_c (ba)=2+\log_a (bc)+ \log_b (ac)+ \log_c (ba)$$

I took LHS and applied base change formula. I changed base to $`\text{abc'}$

Let $abc=\mu$

$$\implies LHS=\dfrac{(\log_\mu a+ \log_\mu b)(\log_\mu c+ \log_\mu b)(\log_\mu a+ \log_\mu c)}{\log_\mu a\times \log_\mu (b)\times \log_\mu (c)} $$

$$= 2+(\log_b a\times \log_c a\times\log_a bc) +(\log_a b\times \log_c b\times\log_b ac) +(\log_b c\times \log_a c\times\log_c ab) $$

This is what i finally get, which is not equal to RHS. what am i doing wrong here?

Thanks a lot.

Answer 1

I think your method can be refined if you avoid using that $\mu$-substitution. Just rewrite the identity as $$\frac{(\log b + \log c)(\log a + \log c)(\log b + \log a)}{\log a \log b \log c}=2+\frac{\log b + \log c}{\log a}+\frac{\log a + \log c}{\log b}+\frac{\log b+\log a}{\log c}$$ Now multiplying both sides by $\log a \log b \log c$ and expanding all the products you'll be done.

Answer 2

Hint: Let $x = \log_ab, y = \log_bc, z=\log_ca \Rightarrow \log_a{bc} = x+\dfrac{1}{z}, \log_b{ca} = y + \dfrac{1}{x}, \log_c{ab} = \log_ca+\log_cb=z+\dfrac{1}{y}, xyz=1$. Can you continue? I am sure it will get to the answer...