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

Note by Reeshabh Ranjan 5 years ago

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

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

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

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

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

\( 81 \) \( > 64 \)

\( \Rightarrow \) \( x = 6 \)

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_Hint: _ \( 32 = 2^5 \).

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

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

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

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

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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## Comments

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

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

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

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

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

\( 81 \) \( > 64 \)

\( \Rightarrow \) \( x = 6 \)

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_Hint: _\( 32 = 2^5 \).Log in to reply

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

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

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

Log in to reply

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

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