To show of 5[SUP]√2 [SUP] >7, logically.

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  • #1
ssd
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Stuck at the simple looking homework problem of my student:
To show 5√2 >7, logically.
 

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  • #2
Delta2
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Its ##5^{\sqrt2}>5^{1.4}##.

Now I suppose you aren't allowed to calculate powers other than integer powers, so you ll have to calculate 5^1.4 using the generalized binomial expansion ##(1+4)^{1.4}##. The first few terms up to 4^3 would be enough to show that ##5^{1.4}>7##.

https://en.wikipedia.org/wiki/Binomial_theorem#Generalisations
 
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  • #3
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Its ##5^{\sqrt2}>5^{1.4}##.

Now I suppose you aren't allowed to calculate powers other than integer powers, so you ll have to calculate 5^1.4 using the generalized binomial expansion ##(1+4)^{1.4}##. The first few terms up to 4^3 would be enough to show that ##5^{1.4}>7##.

https://en.wikipedia.org/wiki/Binomial_theorem#Generalisations
No need for that. If you want to prove ##5^{1.4} > 7##, then this is equivalent to proving ##(5^{14/10})^{10} > 7^{10}##. This should land you in the realm on the integers.
 
  • #4
ssd
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Thanks for your response. So, we can start with 2>1.96 and calculate 514 and 710. But this is not the answer I was looking for. Actually this can be done using standard inequalities (it's my guess only).
 
  • #5
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Thanks for your response. So, we can start with 2>1.96 and calculate 514 and 710. But this is not the answer I was looking for. Actually this can be done using standard inequalities (it's my guess only).
How about letting L = log base 2 and noticing:
L5√2 = √2L5 > 1.4L5 = 4(L5)/10 + L5 = (L54)/10 + L5 > .9 + L5 > ½ + L5 > L(1.4) + L5 = L(1.4•5) = L7.
 

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