Answer :
To simplify the expression [tex]\(\frac{2^{n+2} \cdot 4^{n+1}}{8^{n+1}}\)[/tex], we can use properties of exponents. Let's go step-by-step.
### Step 1: Rewrite all terms with base 2.
First, note that [tex]\(4\)[/tex] and [tex]\(8\)[/tex] can be expressed as powers of [tex]\(2\)[/tex]:
[tex]\[4 = 2^2 \quad \text{and} \quad 8 = 2^3\][/tex]
Thus, [tex]\(4^{n+1}\)[/tex] and [tex]\(8^{n+1}\)[/tex] can be rewritten using base 2:
[tex]\[4^{n+1} = (2^2)^{n+1} = 2^{2(n+1)}\][/tex]
[tex]\[8^{n+1} = (2^3)^{n+1} = 2^{3(n+1)}\][/tex]
### Step 2: Substitute these into the original expression.
This gives us:
[tex]\[ \frac{2^{n+2} \cdot 2^{2(n+1)}}{2^{3(n+1)}} \][/tex]
### Step 3: Simplify the expression using properties of exponents.
Recall that when multiplying like bases, you add the exponents, and when dividing, you subtract the exponents:
[tex]\[ \frac{2^{n+2} \cdot 2^{2(n+1)}}{2^{3(n+1)}} = \frac{2^{n+2 + 2(n+1)}}{2^{3(n+1)}} \][/tex]
### Step 4: Combine exponents in the numerator.
First, simplify the exponents in the numerator:
[tex]\[ n+2 + 2(n+1) = n + 2 + 2n + 2 = 3n + 4 \][/tex]
So the expression becomes:
[tex]\[ \frac{2^{3n+4}}{2^{3(n+1)}} \][/tex]
### Step 5: Simplify the fraction by subtracting the exponents.
Now, subtract the exponent in the denominator from the exponent in the numerator:
[tex]\[ 2^{3n+4 - 3(n+1)} \][/tex]
Simplify the exponent:
[tex]\[ 3n + 4 - 3(n+1) = 3n + 4 - 3n - 3 = 4 - 3 = 1 \][/tex]
So, the original expression simplifies to:
[tex]\[ 2^1 = 2 \][/tex]
### Final Answer
The simplified form of the expression [tex]\(\frac{2^{n+2} \cdot 4^{n+1}}{8^{n+1}}\)[/tex] is:
[tex]\[ \boxed{2} \][/tex]
### Step 1: Rewrite all terms with base 2.
First, note that [tex]\(4\)[/tex] and [tex]\(8\)[/tex] can be expressed as powers of [tex]\(2\)[/tex]:
[tex]\[4 = 2^2 \quad \text{and} \quad 8 = 2^3\][/tex]
Thus, [tex]\(4^{n+1}\)[/tex] and [tex]\(8^{n+1}\)[/tex] can be rewritten using base 2:
[tex]\[4^{n+1} = (2^2)^{n+1} = 2^{2(n+1)}\][/tex]
[tex]\[8^{n+1} = (2^3)^{n+1} = 2^{3(n+1)}\][/tex]
### Step 2: Substitute these into the original expression.
This gives us:
[tex]\[ \frac{2^{n+2} \cdot 2^{2(n+1)}}{2^{3(n+1)}} \][/tex]
### Step 3: Simplify the expression using properties of exponents.
Recall that when multiplying like bases, you add the exponents, and when dividing, you subtract the exponents:
[tex]\[ \frac{2^{n+2} \cdot 2^{2(n+1)}}{2^{3(n+1)}} = \frac{2^{n+2 + 2(n+1)}}{2^{3(n+1)}} \][/tex]
### Step 4: Combine exponents in the numerator.
First, simplify the exponents in the numerator:
[tex]\[ n+2 + 2(n+1) = n + 2 + 2n + 2 = 3n + 4 \][/tex]
So the expression becomes:
[tex]\[ \frac{2^{3n+4}}{2^{3(n+1)}} \][/tex]
### Step 5: Simplify the fraction by subtracting the exponents.
Now, subtract the exponent in the denominator from the exponent in the numerator:
[tex]\[ 2^{3n+4 - 3(n+1)} \][/tex]
Simplify the exponent:
[tex]\[ 3n + 4 - 3(n+1) = 3n + 4 - 3n - 3 = 4 - 3 = 1 \][/tex]
So, the original expression simplifies to:
[tex]\[ 2^1 = 2 \][/tex]
### Final Answer
The simplified form of the expression [tex]\(\frac{2^{n+2} \cdot 4^{n+1}}{8^{n+1}}\)[/tex] is:
[tex]\[ \boxed{2} \][/tex]
