Radical Roots

Algebra Level 3

We know that $2^3 < 3^2$ .
What is the relationship between $2^{1/2}$ and $3^{1/3}$ ?

2^{1/2} < 3^{1/3}

2^{1/2} = 3^{1/3}

2^{1/2} > 3^{1/3}

4 solutions

Seth Christman
Sep 21, 2016

$2^3 < 3^2$

Raise both sides to the $\dfrac{1}{3}$ power.

$2^{3\times{1/3}} < 3^{2\times{1/3}}$

$2 < 3^{2/3}$

Raise both sides to the $\dfrac{1}{2}$ power.

$2^{1/2} < 3^{{2/3}\times{1/2}}$

$2^{1/2} < 3^{1/3}$

Why not raise both sides to the power 1/6 and complete the solution in one step?

Aditya Dhawan - 4 years, 8 months ago

That would have been faster. I felt this way was easier to follow for someone who is completely stuck

Seth Christman - 4 years, 8 months ago

Razzi Masroor
Oct 9, 2016

since 2 cubed is greater than 3 squared,by 6th rooting both sides you get that the cube root of three is more that the square root of 2

Bloons Qoth
Oct 6, 2016

If $2^3 < 3^2$

$\begin{aligned} 2^{\frac{1}{2}} & < 3^{\frac{1}{3}} \\ \big({2^{\frac{1}{2}}}\big)^6 & <\big({2^{\frac{1}{3}}}\big)^6 \\ 2^{\frac{1}{2}*6} & < 3^{\frac{1}{3}*6} \\ 2^3 & <\, 3^2 \end{aligned}$

But it's already given, or obviously, that $2^3 < 3^2$

Therefore the answer is $\color{#69047E}{\boxed{2^{1/2} < 3^{1/3}}}$

Unfortunately, this is not correct. If $x\ne y$ , then $x^6 \ne y^6$ is not necessarily true. For example, take $x=2, y = -2$ .

Pi Han Goh - 4 years, 8 months ago

So how would I write it? Use an inequality sign instead of $\neq ?$

Bloons Qoth - 4 years, 8 months ago

That's a correct approach. Another approach is "If $x>1$ and $y>1$ such that $x > y$ , then $x^n > y^n$ is also true for $n>0$ ."

Pi Han Goh - 4 years, 8 months ago

Daniel Ferreira
Nov 13, 2016

$\mathsf{De \ 2^3 < 3^2, \ temos \ que:}$

$\\ \large\mathsf{2^3 < 3^2} \\\\ \mathsf{\left ( 2^3 \right )^{\frac{1}{6}} < \left ( 3^2 \right )^{\frac{1}{6}}} \\\\ \mathsf{2^{\frac{3}{6}} < 3^{\frac{2}{6}}} \\\\ \boxed{\mathsf{2^{\frac{1}{2}} < 3^{\frac{1}{3}}}}$

Como queríamos demonstrar!

Radical Roots

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