MCQ
Which of the following is the greatest?
- A$7^2$
- ✓$(49)^{3 / 2}$
- C$\left(\frac{1}{343}\right)^{-1 / 3}$
- D$(2401)^{-1 / 4}$
✓
Answer
Correct option: B.
$(49)^{3 / 2}$
(b)
We find that
$(49)^{3 / 2}=\left(7^2\right)^{3 / 2}=7^{2 \times 3 / 2}=7^3$$ \left(\frac{1}{343}\right)^{-1 / 3}=\left(\frac{1}{7^3}\right)^{-1 / 3}=\left(7^{-3}\right)^{-1 / 3}=7^{-3 \times-\frac{1}{3}}=7 $
and $(2401)^{-1 / 4}=\left(7^4\right)^{-1 / 4}=7^{4 \times-\frac{1}{4}}=7^{-1}=\frac{1}{7}$
Clearly, $\frac{1}{7}<7<7^2<7^3 \Rightarrow(2401)^{-1 / 4}<\left(\frac{1}{343}\right)^{-1 / 3}<7^2<(49)^{3 / 2}$ [Using (i), (ii) and (iii)]
Hence, $(49)^{3 / 2}$ is the greatest..
We find that
$(49)^{3 / 2}=\left(7^2\right)^{3 / 2}=7^{2 \times 3 / 2}=7^3$$ \left(\frac{1}{343}\right)^{-1 / 3}=\left(\frac{1}{7^3}\right)^{-1 / 3}=\left(7^{-3}\right)^{-1 / 3}=7^{-3 \times-\frac{1}{3}}=7 $
and $(2401)^{-1 / 4}=\left(7^4\right)^{-1 / 4}=7^{4 \times-\frac{1}{4}}=7^{-1}=\frac{1}{7}$
Clearly, $\frac{1}{7}<7<7^2<7^3 \Rightarrow(2401)^{-1 / 4}<\left(\frac{1}{343}\right)^{-1 / 3}<7^2<(49)^{3 / 2}$ [Using (i), (ii) and (iii)]
Hence, $(49)^{3 / 2}$ is the greatest..
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