Question 120

# The highest number amongst $$\sqrt{2}, \sqrt{3},$$ and $$\sqrt{4}$$ is

Solution

Given that

a = $$\sqrt{2}, b = \sqrt{3},$$ and c = $$\sqrt{4}$$

a = $$(2)^{\frac{1}{2}}$$, b = $$(3)^{\frac{1}{3}}$$ and c = $$(4)^{\frac{1}{4}}$$

Taking log both sides   $$\log_{}{a}$$ = $$\log_{}{(2)^{1/2}}$$ , $$\log_{}{b} = \log_{}{(3)^{1/3}}$$ , $$\log_{}{c} = \log_{}{(4)^{1/4}}$$

we know that ($$\log_{}{2}$$ = 0.3010 , $$\log_{}{3}$$ = 0.4771 , $$\log_{}{4}$$ = $$2\log_{}{2}$$ =0.6020)

Substituting these values

$$\log_{}{a}$$ = $$\frac{1}{2}$$*0.3010 , $$\log_{}{b}$$ =$$\frac{1}{3}$$*0.4771 , $$\log_{}{c}$$= $$\frac{1}{4}$$*0.6020

$$\log_{}{a}$$ = 0.1505 , $$\log_{}{b}$$ = 0.1590 , $$\log_{}{c}$$= 0.1505

We know that if $$A_{1}$$ > $$A_{2}$$ > $$A_{3}$$ > 1   Then  $$log_{}{A_{1}}$$ > $$log_{}{A_{2}}$$ > $$log_{}{A_{3}}$$

Here clearly  $$log_{}{b}$$ > $$log_{}{a}$$ = $$log_{}{c}$$  hence we can say that b = $$\sqrt{3}$$ is the highest number among all.

Alternate method:

a = $$\sqrt{2}, b = \sqrt{3},$$ and c = $$\sqrt{4}$$

$$a^{12} = 2^6, b^{12} = 3^4, c^{12} = 4^3$$

$$a^{12} = 64, b^{12} = 81, c^{12} = 64$$

We can see that $$b^{12}$$ > $$c^{12}$$ = $$a^{12}$$

Also, a, b, c > 1. Hence, we can say that b > a = c.

***  Point to remember ------ $$\sqrt[N]{N}$$ is highest for N=3 (N being natural number) ***

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