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Question Number 11048 by Joel576 last updated on 09/Mar/17

Which one is largest? (without using calculator) 31^(11) or 17^(14) ??

$$\mathrm{Which}\:\mathrm{one}\:\mathrm{is}\:\mathrm{largest}?\:\left(\mathrm{without}\:\mathrm{using}\:\mathrm{calculator}\right) \\ $$$$\mathrm{31}^{\mathrm{11}} \:\mathrm{or}\:\mathrm{17}^{\mathrm{14}} \:\:?? \\ $$

Answered by ajfour last updated on 09/Mar/17

Answered by ajfour last updated on 09/Mar/17

Commented by ajfour last updated on 09/Mar/17

so 17^(14 ) should be greater.

$${so}\:\mathrm{17}^{\mathrm{14}\:} \:{should}\:{be}\:{greater}. \\ $$

Commented by Joel576 last updated on 10/Mar/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by ajfour last updated on 09/Mar/17

log (31^(11)) =11(log 31) =11(1+log 3.1)=11(1+0.48) =11×1.48=14.8+1.48=16.28 while log (17^(14) )=14(log 17) ≈ 14×4(log 2)=56×0.302≈ 17 thus 17^(14 ) should be greater.

$$\mathrm{log}\:\left(\mathrm{31}^{\left.\mathrm{11}\right)} =\mathrm{11}\left(\mathrm{log}\:\:\mathrm{31}\right)\right. \\ $$$$=\mathrm{11}\left(\mathrm{1}+\mathrm{log}\:\mathrm{3}.\mathrm{1}\right)=\mathrm{11}\left(\mathrm{1}+\mathrm{0}.\mathrm{48}\right) \\ $$$$=\mathrm{11}×\mathrm{1}.\mathrm{48}=\mathrm{14}.\mathrm{8}+\mathrm{1}.\mathrm{48}=\mathrm{16}.\mathrm{28} \\ $$$${while}\:\mathrm{log}\:\left(\mathrm{17}^{\mathrm{14}} \right)=\mathrm{14}\left(\mathrm{log}\:\mathrm{17}\right) \\ $$$$\:\approx\:\mathrm{14}×\mathrm{4}\left(\mathrm{log}\:\mathrm{2}\right)=\mathrm{56}×\mathrm{0}.\mathrm{302}\approx\:\mathrm{17} \\ $$$${thus}\:\mathrm{17}^{\mathrm{14}\:} \:{should}\:{be}\:{greater}. \\ $$

Answered by mrW1 last updated on 09/Mar/17

without using calculator: log_2 (31^(11) )=11×log_2 31 <11×log_2 32=11×log_2 2^5 =11×5=55 log_2 (17^(14) )=14×log_2 17 >14×log_2 16=14×log_2 2^4 =14×4=56 ∴ log_2 (17^(14) )>56>55>log_2 (31^(11) ) ∵ 17^(14) >31^(11)

$${without}\:{using}\:{calculator}: \\ $$$$ \\ $$$$\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{31}^{\mathrm{11}} \right)=\mathrm{11}×\mathrm{log}_{\mathrm{2}} \:\mathrm{31} \\ $$$$<\mathrm{11}×\mathrm{log}_{\mathrm{2}} \:\mathrm{32}=\mathrm{11}×\mathrm{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{5}} =\mathrm{11}×\mathrm{5}=\mathrm{55} \\ $$$$ \\ $$$$\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{17}^{\mathrm{14}} \right)=\mathrm{14}×\mathrm{log}_{\mathrm{2}} \:\mathrm{17} \\ $$$$>\mathrm{14}×\mathrm{log}_{\mathrm{2}} \:\mathrm{16}=\mathrm{14}×\mathrm{log}_{\mathrm{2}} \:\mathrm{2}^{\mathrm{4}} =\mathrm{14}×\mathrm{4}=\mathrm{56} \\ $$$$ \\ $$$$\therefore\:\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{17}^{\mathrm{14}} \right)>\mathrm{56}>\mathrm{55}>\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{31}^{\mathrm{11}} \right) \\ $$$$\because\:\mathrm{17}^{\mathrm{14}} >\mathrm{31}^{\mathrm{11}} \\ $$

Commented by Joel576 last updated on 10/Mar/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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