- #1

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Stuck at the simple looking homework problem of my student:

To show 5

To show 5

^{√2}>7, logically.You are using an out of date browser. It may not display this or other websites correctly.

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- Thread starter ssd
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- #1

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Stuck at the simple looking homework problem of my student:

To show 5^{√2} >7, logically.

To show 5

- #2

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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

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.

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- #5

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How about letting L = log base 2 and noticing:^{14}and 7^{10}. But this is not the answer I was looking for. Actually this can be done using standard inequalities (it's my guess only).

L5

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