If \( n \) and \( k \) are numbers greater than 1 and \( \sqrt[4]{n^{5}} \) is equivalent to \( \sqrt[3]{k^{2}} \), for what value of \( a \) is \( n^{2 a+1} \) equal to \( k \) ?

We start with the given equation: \[ \sqrt[4]{n^5} = \sqrt[3]{k^2} \] To express both sides as powers, we can rewrite them: \[ n^{5/4} = k^{2/3} \] Next, we want to express \( k \) in terms of \( n \). To do that, we can raise both sides to the common power that eliminates the fractions. The least common multiple of the denominators 4 and 3 is 12, so we raise both sides to the power of 12: \[ (n^{5/4})^{12} = (k^{2/3})^{12} \] This simplifies to: \[ n^{15} = k^8 \] We can express \( k \) in terms of \( n \): \[ k = n^{15/8} \] The problem asks for the value of \( a \) such that \( n^{2a+1} = k \). We substitute \( k \) from above: \[ n^{2a+1} = n^{15/8} \] Since the bases are the same, we can equate the exponents: \[ 2a + 1 = \frac{15}{8} \] Now, we solve for \( a \). First, we isolate \( 2a \): \[ 2a = \frac{15}{8} - 1 \] Converting 1 to a fraction with a denominator of 8 gives us: \[ 1 = \frac{8}{8} \] Thus, \[ 2a = \frac{15}{8} - \frac{8}{8} = \frac{7}{8} \] Now we divide by 2 to find \( a \): \[ a = \frac{7/8}{2} = \frac{7}{16} \] Thus, the value of \( a \) is: \[ \boxed{\frac{7}{16}} \]