Question

If $a^x=b,b^y=c,c^z=a$ and $x={\log }_b{a}^2;y={\log }_c{b}^3\&z={\log }_a{c}^k$ where $a,bc>0\&a,b,c\ne 1$, then $k$ is equal to

• A. $\frac{1}{5}$
• B. $\frac{1}{6}$
• C. ${\log }_{64}2$
• D. ${\log }_{32}2$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{1}{6}$
C ${\log }_{64}2$

Given $a^x=b$
Taking logarithm (w.r.t base $a$) on both sides
$\Rightarrow x={\log }_{a}b$ ....(1)
$b^y=c$
$\Rightarrow y={\log }_{b}c$ .....(2)
$c^z=a$
$\Rightarrow z={\log }_{c}a$ .....(3)
Multiplying these eqns, we get
$xyz=\frac{{\log }_{10}b}{{\log }_{10}a}\frac{{\log }_{10}c}{{\log }_{10}b}\frac{{\log }_{10}a}{{\log }_{10}c}[\because {\log }_{a}b=\frac{\log b}{\log a}]$
$xyz=1$ ...(4)
Also, given $x={\log }_b{a}^2;y={\log }_c{b}^3\&z={\log }_a{c}^k$
$\Rightarrow x=2{\log }_{b}a;y=3{\log }_{c}b\&z=k{\log }_{a}c$
Put these values in (4)
$6k\frac{{\log }_{10}a}{{\log }_{10}b}\frac{{\log }_{10}b}{{\log }_{10}c}\frac{{\log }_{10}c}{{\log }_{10}a}=1$
$\Rightarrow k=\frac{1}{6}$
Also, $k=\frac{{\log }_{10}2}{6{\log }_{10}2}$
$\Rightarrow k=\frac{{\log }_{10}2}{{\log }_{10}{2}^6}$
$\Rightarrow k=\frac{{\log }_{10}2}{{\log }_{10}64}$
$\Rightarrow k={\log }_{64}2$