Which is equivalent to 64 superscript one fourth

Explanation:

One should remember that finding a fourth root is the same as finding the square root twice.

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Algebra Examples

#root(4)x =sqrt(sqrtx)#

#root(4)64 = sqrt(sqrt64) = sqrt8#

However 8 is not a perfect square so it does not have an exact square root.

#sqrt8 = sqrt(4 xx 2) = 2sqrt2#

The expression for which you want to find an equivalent form is:

{(\sqrt[4]{9})}^{\frac{1}{2}x}

Answer:

Here there are 3 equivalent expressions:

{(\sqrt[4]{3^2})}^{\frac{1}{2}x}

(\sqrt{3})^{\frac{1}{2}x}

(\sqrt[4]{3} )^x

Explanation:

There are many equivalent forms that you can find, so I will show some of the most important.

The first step that you should do is to put the radicand number (the number inside the root, i.e. 3) as a product of its prime factors. Thus the first equivalent expression is:

{(\sqrt[4]{3^2})}^{\frac{1}{2}x}

Now you can simplify the root index, 4, with the exponent of the radicand, 2, an you get a new equivalent:

{(\sqrt[4]{3^2})}^{\frac{1}{2}x}=(\sqrt{3})^{\frac{1}{2}x}

Another equivalent form is obtained if you convert the 1/2 index (before the x) into a root index:

(\sqrt{3})^{\frac{1}{2}x}=(\sqrt{\sqrt{3}})^x=(\sqrt[4]{3} )^x

The main properties used to find those equivalent expressions are:

a^{\frac{1}{n}}=\sqrt[n]{a}

And:

{(a^{\frac{1}{n}})^m=a^{\frac{m}{n}}= \sqrt[n]{a^m}

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Gauthmathier7974

Grade 12 · 2021-07-10

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Gauthmathier2905

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