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]
