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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 :(

  • What you've written in your next-to-last line is correct for $ \ x + y + z \ \ , $ rather than the product. Compare that with $$ xyz \ = \ \ \frac{(\log b+\log c)·(\log a + \log c)·(\log a + \log b)}{\log a· \log b · \log c} \ \ . $$ The algebra for $ \ xyz \ - \ [x + y + z] \ $ is a little tedious, but gets you to the correct result. –  May 28 '22 at 08:55

1 Answers1

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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 $\log(p)\times\log(q) \ne \log(pq)$

19aksh
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