Question

Math Riddle

Answer 1

WITHOUT using a calculator, prove that 3^(1/3) is greater than 10^(1/10)

Answer 2

oh god why

Answer 3

looks like the general notation is something \[x^\frac{ 1 }{ x }\]

\[\lim_{x \rightarrow 3}x^\frac{ 1 }{ x }\]
\[\lim_{x \rightarrow 3}\frac{ 1 }{ x }lnx\]
\[\lim_{x \rightarrow 3}\frac{ 1 }{ x }=\frac{ 1 }{ 3 }=e^\frac{ 1 }{ 3 }\]
repeating the limit as x approaches 10
\[\lim_{x \rightarrow 10}=e^\frac{ 1 }{ 10 }\]
\[\huge e^\frac{ 1 }{ 3 }\ge e^\frac{ 1 }{ 10 }\]
no calculator used (not sure if correct haha)

Answer 4

Also i can explain my steps a little more but i assumed some knowledge of calculus (L'hopital's rule) is known.

Answer 5

awesome, thanks

Answer 6

there's nothing indeterminate here so wondering how does L'H apply ?

this is boring approach, but i think it works:


\(\dfrac{3^{{10\over 3}}}{10} = \dfrac{3^{{9\over 3}} 3^{{1\over 3}}}{10} = 2.7 \times 3^{{1\over 3}} \)

Clearly, \( 3^{{1\over 3}} > 1\), ie if \(a \cdot a \cdot a = 3\) then \(a >1\). So \(\dfrac{3^{{10\over 3}}}{10} > 1 \)

Now, if \( \dfrac{3^{{10\over 3}}}{10} > 1\), ie \(\left( \dfrac{3^{{1\over 3}}}{10^{{1\over 10}}} \right)^{10} > 1 \), then \(\dfrac{3^{ {1 \over 3}}} {10^{{1 \over 10}}} > 1\) and so \(3^{ {1 \over 3}} > 10^{{1 \over 10}} \)