Question 77

The smallest number among
$$\sqrt[3]{4}, \sqrt[5]{7}, \sqrt[4]{5}, \sqrt{3}$$ is


Solution :

we need to find the LCM of roots' power i.e. LCM of 3,5,4,2 i.e. 60

$$\sqrt[3]{4}$$ =$$4^{\frac{1}{3}}\ =\ 4^{\left(\frac{20}{60}\right)}$$  =  $$\left(4^{50}\right)^{\frac{1}{60}}$$

$$\sqrt[5]{7}$$ =$$7^{\frac{1}{5}}\ =\ 7^{\left(\frac{12}{60}\right)}$$  = $$\left(7^{12}\right)^{\frac{1}{60}}$$

$$\sqrt[4]{5}$$ =$$5^{\frac{1}{4}}\ =\ 5^{\left(\frac{15}{60}\right)}$$  = $$\left(5^{15}\right)^{\frac{1}{60}}$$

 $$\sqrt[2]{3}$$ =$$3^{\frac{1}{2}}\ =\ 3^{\left(\frac{30}{60}\right)}$$ = $$\left(3^{30}\right)^{\frac{1}{60}}$$

As all the external powers are same, we just need to arrange the internal powers in ascending order to find the lowest one.

On careful observation, and also relying on the fact that 7 has the lowest power.  $$\sqrt[5]{7}$$  is the lowest one.Answer.

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