Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Let's start by examining the given equation:
[tex]\[ \left(2^1\right)^n = \frac{2^x}{8^y} \][/tex]
First, simplify the left-hand side of the equation. Since [tex]\(\left(2^1\right)^n\)[/tex] is simply [tex]\(2^n\)[/tex], the equation becomes:
[tex]\[ 2^n = \frac{2^x}{8^y} \][/tex]
Next, simplify the right-hand side. Notice that [tex]\(8\)[/tex] can be expressed as a power of [tex]\(2\)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
Therefore, [tex]\(8^y\)[/tex] can be expressed as:
[tex]\[ 8^y = (2^3)^y = 2^{3y} \][/tex]
So, the equation now is:
[tex]\[ 2^n = \frac{2^x}{2^{3y}} \][/tex]
When dividing exponential expressions with the same base, you subtract the exponents:
[tex]\[ \frac{2^x}{2^{3y}} = 2^{x - 3y} \][/tex]
Now, the equation is:
[tex]\[ 2^n = 2^{x - 3y} \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ n = x - 3y \][/tex]
Thus, expressing [tex]\(n\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex], we have:
[tex]\[ n = x - 3y \][/tex]
[tex]\[ \left(2^1\right)^n = \frac{2^x}{8^y} \][/tex]
First, simplify the left-hand side of the equation. Since [tex]\(\left(2^1\right)^n\)[/tex] is simply [tex]\(2^n\)[/tex], the equation becomes:
[tex]\[ 2^n = \frac{2^x}{8^y} \][/tex]
Next, simplify the right-hand side. Notice that [tex]\(8\)[/tex] can be expressed as a power of [tex]\(2\)[/tex]:
[tex]\[ 8 = 2^3 \][/tex]
Therefore, [tex]\(8^y\)[/tex] can be expressed as:
[tex]\[ 8^y = (2^3)^y = 2^{3y} \][/tex]
So, the equation now is:
[tex]\[ 2^n = \frac{2^x}{2^{3y}} \][/tex]
When dividing exponential expressions with the same base, you subtract the exponents:
[tex]\[ \frac{2^x}{2^{3y}} = 2^{x - 3y} \][/tex]
Now, the equation is:
[tex]\[ 2^n = 2^{x - 3y} \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ n = x - 3y \][/tex]
Thus, expressing [tex]\(n\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex], we have:
[tex]\[ n = x - 3y \][/tex]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.