Question

Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

Answer 1

To solve the problem we must know the Basic Rules of Exponentiation.

Basic Rules of Exponentiation

• $x^ax^b = x^{(a+b)}$
• $\dfrac{x^a}{x^b} = x^{(a-b)}$
• $(a^a)^b =x^{(a\times b)}$
• $(xy)^a = x^ay^a$

• $x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4$

The solution of the expression is $\dfrac{4x^4}{y^6}$.

Explanation

Given to us

• $(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

$=2^2x^4y^{-6}$

Using identity $\dfrac{x^a}{x^b} = x^{(a-b)}$,

$=\dfrac{2^2x^4}{y^6}$

$=\dfrac{4x^4}{y^6}$

Hence, the solution of the expression is $\dfrac{4x^4}{y^6}$.

Learn more about Exponentiation:

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