Question
5. \( 2^{-2 n}\left(2^{n+1}\right)=2^{-2 n} \)
Ask by Ayala Simmons. in the United States
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
\( n = -1 \)
Solution
Solve the equation \( 2^{-2n}(2^{n+1})=2^{-2n} \).
Solve the equation by following steps:
- step0: Solve for \(n\):
\(2^{-2n}\times 2^{n+1}=2^{-2n}\)
- step1: Multiply the terms:
\(2^{-n+1}=2^{-2n}\)
- step2: Set the exponents equal:
\(-n+1=-2n\)
- step3: Move the variable to the left side:
\(-n+1+2n=0\)
- step4: Add the terms:
\(n+1=0\)
- step5: Move the constant to the right side:
\(n=0-1\)
- step6: Remove 0:
\(n=-1\)
The solution to the equation \(2^{-2n}(2^{n+1})=2^{-2n}\) is \(n=-1\).
Reviewed and approved by the UpStudy tutoring team
Like
error msg
Explain 
Simplify this solution Bonus Knowledge
To solve the equation \( 2^{-2 n}(2^{n+1})=2^{-2 n} \), start by simplifying the left side. You have \( 2^{-2 n} \cdot 2^{n + 1} \). When multiplying exponents with the same base, you can add the exponents: \[ 2^{-2n + (n + 1)} = 2^{-2n + n + 1} = 2^{-n + 1}. \] So the equation now becomes: \[ 2^{-n + 1} = 2^{-2n}. \] Since the bases are the same, you can set the exponents equal to each other: \[ -n + 1 = -2n. \] Now, isolate \(n\): \[ 1 = -2n + n, \] \[ 1 = -n, \] \[ n = -1. \] Thus, the solution to the equation is \( n = -1 \).
