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The greatest integer x which satisfy $$3^{20} > 32^{x}$$.

Note by Reeshabh Ranjan
4 years, 4 months ago

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We have to find the greatest integer which satisfies $$3^{20}$$ $$> 32^x$$

$$\Rightarrow$$ $$3^{20}$$ $$> 2^{5x}$$

$$\Rightarrow$$ $$\sqrt[5]{3^{20}}$$ $$> \sqrt[5]{2^{5x}}$$

$$\Rightarrow$$ $$3^4$$ $$> 2^x$$

$$\Rightarrow$$ $$81$$ $$> 2^x$$

$$81$$ $$> 64$$

$$\Rightarrow$$ $$x = 6$$

- 4 years, 4 months ago

_Hint: _ $$32 = 2^5$$.

Staff - 4 years, 4 months ago

step 1 :: 20 log3 > x log 32. step 2 :: 20*0.477 > x 1.505. step 3 :: 6.33 > x. therefore x = 6.

- 4 years, 4 months ago

do reeshab u read at fiitjee cuz this problem was given there

- 4 years, 4 months ago

hey! are you in fiitjee? if yes please tell the centre. I am in fiitjee south delhi.

- 4 years, 4 months ago

You all are awesome, Brilliant.org is so cool :D

- 3 years, 9 months ago