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Question Number 196582 by MATHEMATICSAM last updated on 27/Aug/23

$$\mathrm{If}x={\log }_{a}bc,y={\log }_{b}ca\mathrm{and}z={\log }_{c}ab$$$$\mathrm{then}\mathrm{prove}\mathrm{that}x+y+z=xyz-2.$$

Answered by som(math1967) last updated on 27/Aug/23

$$x={log}_{a}bc\Rightarrow 1+x={log}_{a}a+{log}_{a}bc$$$$\Rightarrow 1+x={log}_{a}abc\Rightarrow \frac{1}{1+x}={log}_{abc}a$$$$sameway\frac{1}{1+y}={log}_{abc}b$$$$\frac{1}{1+z}={log}_{abc}c$$$$\therefore \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}={log}_{abc}abc=1$$$$\Rightarrow \frac{1+y+1+x}{(1+x)(1+y)}=\frac{z}{1+z}$$$$\Rightarrow 2z+yz+zx+2+x+y=z(1+x+y+xy)$$$$\Rightarrow 2z+yz+zx+2+x+y$$$$=z+zx+yz+xyz$$$$\therefore x+y+z=xyz-2$$

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