Explore IDNLearn.com's extensive Q&A database and find the answers you need. Get accurate and detailed answers to your questions from our knowledgeable and dedicated community members.
Sagot :
To solve the equation [tex]\(\left(x^2\right)^y = x^{16}\)[/tex], we need to use the properties of exponents. Let's proceed step-by-step.
1. Rewrite the left-hand side using the exponentiation rule: [tex]\((a^m)^n = a^{m \cdot n}\)[/tex]. Therefore, [tex]\(\left(x^2\right)^y\)[/tex] becomes:
[tex]\[ (x^2)^y = x^{2y} \][/tex]
2. Equate the exponents on both sides of the equation since the bases on both sides are the same (both are [tex]\(x\)[/tex]):
[tex]\[ x^{2y} = x^{16} \][/tex]
3. Solve for [tex]\(y\)[/tex] by setting the exponents equal to each other:
[tex]\[ 2y = 16 \][/tex]
4. Divide both sides by 2:
[tex]\[ y = \frac{16}{2} \][/tex]
5. Simplify the right-hand side:
[tex]\[ y = 8 \][/tex]
So, [tex]\(y = 8\)[/tex].
Given the choices:
A) 4
B) 6
C) 8
D) 14
The correct equivalent value of [tex]\(y\)[/tex] is [tex]\(8\)[/tex], which corresponds to option C.
1. Rewrite the left-hand side using the exponentiation rule: [tex]\((a^m)^n = a^{m \cdot n}\)[/tex]. Therefore, [tex]\(\left(x^2\right)^y\)[/tex] becomes:
[tex]\[ (x^2)^y = x^{2y} \][/tex]
2. Equate the exponents on both sides of the equation since the bases on both sides are the same (both are [tex]\(x\)[/tex]):
[tex]\[ x^{2y} = x^{16} \][/tex]
3. Solve for [tex]\(y\)[/tex] by setting the exponents equal to each other:
[tex]\[ 2y = 16 \][/tex]
4. Divide both sides by 2:
[tex]\[ y = \frac{16}{2} \][/tex]
5. Simplify the right-hand side:
[tex]\[ y = 8 \][/tex]
So, [tex]\(y = 8\)[/tex].
Given the choices:
A) 4
B) 6
C) 8
D) 14
The correct equivalent value of [tex]\(y\)[/tex] is [tex]\(8\)[/tex], which corresponds to option C.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is your go-to source for accurate answers. Thanks for stopping by, and come back for more helpful information.