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 January 21st, 2014, 03:28 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 which largest which largest with out calculator 31^11 or 17^14
 January 21st, 2014, 10:01 AM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 634 Thanks: 96 Math Focus: Electrical Engineering Applications Re: which largest Hi mared, $31^{11} \lt 32^{11}, \ 32^{11}=\left(2^5\right)^{11}$ $16^{14} \lt 17^{14}, \ 16^{14}= \left(2^4\right)^{14}$ Take $\ \log_2 \text{ of } \left(2^5\right)^{11} \text{ and } \left(2^4\right)^{14}$ $\log_2\left(2^5\right)^{11}=11 \cdot 5 =55$ $\log_2 \left(2^4\right)^{14}=14 \cdot 4=56$ $32^{11} \lt 16^{14} \$ so $31^{11} \lt 17^{14}$
January 21st, 2014, 01:27 PM   #3
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Re: which largest

Hello, mared!

A no-logs approach . . .

Quote:
 $\text{Which is larger (without calculator)? }\; 31^{11}\,\text{ or }\,17^{14}$

$31^{11} \;=<\;(32)^{11} \;=\;(2^5)^{11} \;\;\;\Rightarrow\;\;\;31^{11}\;<\;2^{55}=$

$17^{14} \;=>\;16^{14} \;=\;(2^4)^{14} \;\;\;\Rightarrow\;\;\;17^{14}\;>\;2^{56}=$

$\text{W\!e have: }\:17^{14} \;\>\:2^{56} \;>\;2^{55} \;>\;31^{11}$

$\text{Therefore, }17^{14}\text{ is larger.}$

 January 21st, 2014, 08:16 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,307 Thanks: 1976 Now prove that 17^14 > 6(31^11).

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