Certainly! Let's solve the given equation step-by-step:
Given equation:
[tex]\[
2^{2y} \times 2^{3y + 2} = \frac{8^{5y}}{4^n}
\][/tex]
Step 1: Simplify the left-hand side (LHS)
We use the property of exponents that states [tex]\(a^m \times a^n = a^{m+n}\)[/tex]:
[tex]\[
2^{2y} \times 2^{3y + 2} = 2^{2y + (3y + 2)} = 2^{5y + 2}
\][/tex]
Step 2: Simplify the right-hand side (RHS)
First, express the terms in the denominator and numerator with the same base, which is base 2.
[tex]\[
8^{5y} = (2^3)^{5y} = 2^{3 \cdot 5y} = 2^{15y}
\][/tex]
[tex]\[
4^n = (2^2)^n = 2^{2n}
\][/tex]
Now combine these results in the fraction:
[tex]\[
\frac{8^{5y}}{4^n} = \frac{2^{15y}}{2^{2n}} = 2^{15y - 2n}
\][/tex]
Step 3: Combine and equate exponents
Having simplified both sides, we have:
[tex]\[
2^{5y + 2} = 2^{15y - 2n}
\][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[
5y + 2 = 15y - 2n
\][/tex]
Step 4: Solve for [tex]\(n\)[/tex] in terms of [tex]\(y\)[/tex]
Rearrange the equation to isolate [tex]\(n\)[/tex]:
[tex]\[
5y + 2 = 15y - 2n
\][/tex]
Subtract [tex]\(5y\)[/tex] from both sides:
[tex]\[
2 = 10y - 2n
\][/tex]
Add [tex]\(2n\)[/tex] to both sides to isolate terms involving [tex]\(n\)[/tex]:
[tex]\[
2 + 2n = 10y
\][/tex]
Now, subtract 2 from both sides:
[tex]\[
2n = 10y - 2
\][/tex]
Finally, divide by 2 to solve for [tex]\(n\)[/tex]:
[tex]\[
n = \frac{10y - 2}{2} = 5y - 1
\][/tex]
Thus, the expression for [tex]\(n\)[/tex] in terms of [tex]\(y\)[/tex] is:
[tex]\[
n = 5y - 1
\][/tex]
This is the simplified expression showing [tex]\(n\)[/tex] in terms of [tex]\(y\)[/tex].
