Question

Assume that all variables are positive.
Part A: Show that .
Part B: Simplify .
Part C: If , what is the value of h in terms of g?
(going to post equations soon)

Answer 1

\[part A:\sqrt{\sqrt{x}}=\sqrt[4]{x}\]\[part B:\sqrt[4]{\sqrt[4]{x}}\]\[part C: \sqrt[g]{\sqrt[g]{x}}=\sqrt[h]{x}\]

Answer 2

i wonder what "show" means here...

Answer 3

if you use exponents you get it, but that is just a trick of notation

Answer 4

\[\sqrt{\sqrt{x}}=\left(x^{\frac{1}{2}}\right)^{\frac{1}{2}}\]
\[=x^{\frac{1}{2}\times \frac{1}{2}}\]
\[=x^{\frac{1}{4}}\]
\[=\sqrt[4]{x}\]

Answer 5

same for B
\[\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\] giving you
\[\sqrt[16]{x}\]

Answer 6

\[\sqrt{x}=x ^{1/2}~~~~~so~~~~\sqrt{\sqrt{x}}=[x ^{1/2}]^{1/2} = x ^{1/2 * 1/2} = x ^{1/4}\]

Answer 7

lol you beat me to it.

Answer 8

and for the last one, just like the first two
\[g^2=h\]

Answer 9

I think the point of this is to realize the property

\[\sqrt[n]{x} = x ^{1/n}\]

Answer 10

then the normal 3 or 4 exponent properties

Answer 11

\[\sqrt[g]{\sqrt[g]{x}} = [\sqrt[g]{x}]^{1/g} = [x ^{1/g}]^{1/g} = x ^{1/g^2}~~~~~and\]
\[\sqrt[h]{x} = x ^{1/h}\]
setting equal
\[x ^{1/g^2} = x ^{1/h}~~~~so~~~~\frac{ 1 }{ h^2 } = \frac{ 1 }{ g }\]

Answer 12

sorry the h and g are mistyped in the very last expression... it is
\[\frac{ 1 }{ g^2 }=\frac{ 1 }{ h }~~~~so~~~~g^2 = h\]