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Please help me out of this mathematics problem.

The greatest integer x which satisfy \(3^{20} > 32^{x}\).

Note by Reeshabh Ranjan
4 years, 2 months ago

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

\( \Rightarrow \) \( 3^{20} \) \( > 2^{5x} \)

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

\( \Rightarrow \) \( 3^4 \) \( > 2^x \)

\( \Rightarrow \) \( 81 \) \( > 2^x \)

\( 81 \) \( > 64 \)

\( \Rightarrow \) \( x = 6 \) Anoopam Mishra · 4 years, 2 months ago

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_Hint: _ \( 32 = 2^5 \). Calvin Lin Staff · 4 years, 2 months ago

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step 1 :: 20 log3 > x log 32. step 2 :: 20*0.477 > x 1.505. step 3 :: 6.33 > x. therefore x = 6. Satish Kumar · 4 years, 1 month ago

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do reeshab u read at fiitjee cuz this problem was given there Superman Son · 4 years, 2 months ago

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@Superman Son hey! are you in fiitjee? if yes please tell the centre. I am in fiitjee south delhi. Anoopam Mishra · 4 years, 2 months ago

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You all are awesome, Brilliant.org is so cool :D Reeshabh Ranjan · 3 years, 7 months ago

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