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

[ssd]

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

 

[Delta2]

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

 

[micromass]

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.

 

[ssd]

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).

 

[Zafa Pi]

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.