Question

If $x={\log }_{a}bc,y={\log }_b,z={\log }_{c}ab$, then

• A. $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$
• B. $\frac{1}{x-1}+\frac{1}{y-1}+\frac{1}{z-1}=1$
• C. $xyz=x+y+z+1$
• D. $xyz=1$

Answer 1

The correct option is A $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=1$

$\begin{array}{cc}x={\log }_{a}bc& 1+\frac{\log bc}{\log a}=x+1\\ y={\log }_{b}ca& 1+\frac{\log ca}{\log b}=y+1\\ & 1+\frac{\log ab}{\log c}=z+1\\ \Rightarrow \frac{\log a+\log bc}{\log a}=x+1& \\ \frac{\log b+\log ca}{\log b}=y+1& \\ \frac{\log c+\log ab}{\log c}=z+1& \\ \Rightarrow \frac{\log abc}{\log a}=x+1& \\ \frac{\log abc}{\log b}=y+1& \\ \frac{\log abc}{\log c}=z+1& \\ \Rightarrow \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=\frac{\log a}{\log abc}+\frac{\log b}{\log abc}+\frac{\log c}{\log abc}& \\ \Rightarrow \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=\frac{\log a+\log b+\log c}{\log abc}& \\ \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=1& \end{array}$

Hence, this is the answer.