Use the like bases property to solve the equation \( \left(\frac{1}{2}\right)^{n+1}=2^{5 n-5} \) \( n=\square \)

First, let's rewrite \( 2^{5n-5} \) using the property of exponents. Since \( \frac{1}{2} = 2^{-1} \), we can express \( \left(\frac{1}{2}\right)^{n+1} \) as: \[ \left(2^{-1}\right)^{n+1} = 2^{-(n+1)} = 2^{-n-1} \] Now our equation becomes: \[ 2^{-n-1} = 2^{5n-5} \] Since the bases are the same, we can set the exponents equal to each other: \[ -n - 1 = 5n - 5 \] Next, we can solve for \( n \). First, add \( n \) to both sides: \[ -1 = 6n - 5 \] Now, add 5 to both sides: \[ 4 = 6n \] Finally, divide both sides by 6: \[ n = \frac{4}{6} = \frac{2}{3} \] So, the solution is: \[ n=\frac{2}{3} \]