Question

Simplify this $ \sqrt {{9^{16{x^2}}}} $ ?

Answer 1

Hint: In order to simplify the given value, we can write the square root given in terms of power which is $ \dfrac{1}{2} $ , and the value becomes $ {\left( {{9^{16{x^2}}}} \right)^{\dfrac{1}{2}}} $ . Then by using the law of radicals multiplying the powers to get our results. We can also further simplify it or can leave as it is.

Complete step-by-step answer:
We are given with the value $ \sqrt {{9^{16{x^2}}}} $ .
Since, we know that the square can be written in the terms as the power of $ \dfrac{1}{2} $ , and by this we get that: $ \sqrt {{9^{16{x^2}}}} = {\left( {{9^{16{x^2}}}} \right)^{\dfrac{1}{2}}} $ .
From the Law of radicals, we know that $ {\left( {{a^p}} \right)^q} = {a^{p.q}} $ .
Using this in our equation we get that:
 $ {\left( {{9^{16{x^2}}}} \right)^{\dfrac{1}{2}}} = {9^{16{x^2}}}^{ \times \dfrac{1}{2}} = {9^{8{x^2}}} $
Since, we know that $ {\left( 3 \right)^2} = 9 $ . So, we can write our equation as:
 $ {9^{8{x^2}}} = {\left( {{3^2}} \right)^{8{x^2}}} $ .
Again, implementing the Law of radicals and we get:
 $ {9^{8{x^2}}} = {\left( {{3^2}} \right)^{8{x^2}}} = {3^{16{x^2}}} $ .
Hence, the value of $ \sqrt {{9^{16{x^2}}}} = {3^{16{x^2}}} $

Alternate method:
We are given with the value: $ \sqrt {{9^{16{x^2}}}} $
Let the value $ {9^{16{x^2}}} $ be $ {a^{2m}} $ .
As, we know that $ {a^{2m}} $ can be written as $ {\left( {{a^m}} \right)^2} $ .
So, we can write our value as $ \sqrt {{9^{16{x^2}}}} = \sqrt {{9^{2 \times 8{x^2}}}} = \sqrt {{{\left( {{9^{8{x^2}}}} \right)}^2}} $ .
Since, we know that the square root of a square is only the value, that is $ \sqrt {{a^2}} = a $ .
Applying this in the above equation, we get:
 $ \sqrt {{{\left( {{9^{8{x^2}}}} \right)}^2}} = {9^{8{x^2}}} $ .
We can further solve it to get more simplest terms or else can leave it at this place only.
Hence, $ \sqrt {{9^{16{x^2}}}} $ is equal to $ {9^{8{x^2}}} $ .
So, the correct answer is “ $ {9^{8{x^2}}} $ ”.

Note: $ {\left( {{a^p}} \right)^q} = {a^{p.q}} $
When the base would be the same during multiplication then the powers would be added to the variables and would be stored in any one variable present. For ex: $ {p^a}.{p^b} = {p^{a + b}} $
When the base would be the same during division of variables, then powers would be subtracted. For Ex: $ \dfrac{{{p^a}}}{{{p^b}}} = {p^{a - b}} $