If $q^{-a} = \frac{1}{r}$ and $r^{-b} = \frac{1}{s}$ and $s^{-c} = \frac{1}{q}$, the value of $abc$ is _____.

$(rqs)^{-1}$
0
1
r+q+s

Correct answer: 1

Solution

From $q^{-a}=1/r$, we get $r=q^a$. From $r^{-b}=1/s$, we get $s=r^b=q^{ab}$. From $s^{-c}=1/q$, we get $q=s^c=q^{abc}$. Hence $abc=1$.

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