which-is-greater-4-2-or-3-3-

Question Number 176141 by Linton last updated on 13/Sep/22

w h i c h i s g r e a t e r ?

4^{\sqrt{2}} o r 3^{\sqrt{3}}

Commented by Linton last updated on 13/Sep/22

w i t h o u t c a l c u l a t o r s h o w w o r k i n g

Answered by Frix last updated on 13/Sep/22

we know x > 1 \land y > 1 \land z > 1 \land x ≷ y \Rightarrow x^{z} ≷ y^{z}

we also know 4^{\sqrt{2}} > 1 \land 3^{\sqrt{3}} > 1 \land \sqrt{2} > 1

4^{\sqrt{2}} \overset{?}{≷} 3^{\sqrt{3}}

{(4^{\sqrt{2}})}^{\sqrt{2}} \overset{?}{≷} {(3^{\sqrt{3}})}^{\sqrt{2}}

16 \overset{?}{≷} 3^{\sqrt{6}}

we know \sqrt{6} < \frac{5}{2} because {(\frac{5}{2})}^{2} = \frac{25}{4} > 6

3^{\frac{5}{2}} = 9 \sqrt{3}

we know \sqrt{3} < \frac{7}{4} because {(\frac{7}{4})}^{2} = \frac{49}{16} > 3

but 9 \times \frac{7}{4} = \frac{63}{4} < 16

\Rightarrow

16 > 9 \times \frac{7}{4} > 9 \sqrt{3} > 3^{\sqrt{6}}

16 > 3^{\sqrt{6}}

4^{\sqrt{2}} > 3^{\sqrt{3}}

Commented by Linton last updated on 14/Sep/22

w e l l d o n e

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