If $x = \log_a(bc), y = \log_b(ca)$ and $z = \log_c(ab)$, then find the value of $xyz – x- y – z$
This question is given in my book but I'm unable to solve it.
Here's what I got so far:
$$x = \log_a(bc) \implies x + 1 = \log_a(bc) + \log_a(a)\implies \boxed{x + 1= \log_a(abc)}$$
$$y = \log_b(ac) \implies y + 1 = \log_b(ac) + \log_b(b)\implies \boxed{y + 1= \log_b(abc)}$$
$$z = \log_c(ab) \implies z + 1 = \log_c(ab) + \log_c(c)\implies \boxed{z + 1= \log_c(abc)}$$
Now,
$$\dfrac{1}{x+1} = \log_{abc}(a)$$
$$\dfrac{1}{y+1} = \log_{abc}(b)$$
$$\dfrac{1}{z+1} = \log_{abc}(c)$$
So,
$$\boxed{\dfrac{1}{x+1} + \dfrac{1}{y+1} + \dfrac{1}{z+1} = 1}$$
But this relation doesn't seem to help here.
I also tried to find $xyz$,
$$\begin{align}xyz& = \log_a(bc) \cdot \log_b(ac) \cdot \log_c(ab)\\\\& = \dfrac{\log b + \log c}{
\log a}+ \dfrac{\log a + \log c}{
\log b} + \dfrac{\log a + \log b}{
\log c}\\\\& = \dfrac{\log b\cdot \log c(\log b + \log c) + \log a\cdot \log c(\log a + \log c) + \log a\cdot \log b(\log a + \log b)}{
\log a\cdot \log b \cdot \log c}\\\\& = \dfrac{\log b\cdot \log c(\log bc) + \log a\cdot \log c(\log ac) + \log a\cdot \log b(\log ab)}{
\log a\cdot \log b \cdot \log c}\end{align}$$
I messed up 🙁
Best Answer
Your first method will solve it.
$\begin{align}&\frac1{x+1}+\frac1{y+1}+\frac1{z+1} = 1\\\\ \Rightarrow&(y+1)(z+1)+(x+1)(z+1) + (x+1)(y+1) = (x+1)(y+1)(z+1)\\\\ \Rightarrow& xy + yz + zy + 2(x+y+z) +3= xyz + xy + yz +zx + x+y+z+ 1 \\\\\Rightarrow&\boxed{xyz-x-y-z = 2}\end{align}$
In your second method, note that $\log(p)\times\log(q) \ne \log(pq)$