If n and k are numbers greater than 1 and sqrt [2](n^5) is equivalent to sqrt [3](k^4) for what the value of a if n^4a-1 equal tok?

Let the given equation be<br />$$\sqrt[2]{n^5} = \sqrt[3]{k^4}$$<br />$$n^{5/2} = k^{4/3}$$<br />We are given that $n^{4a-1} = k$. We want to find the value of $a$.<br /><br />Raise both sides of $n^{5/2} = k^{4/3}$ to the power of $\frac{6}{5}$:<br />$$(n^{5/2})^{\frac{6}{5}} = (k^{4/3})^{\frac{6}{5}}$$<br />$$n^{\frac{5}{2} \times \frac{6}{5}} = k^{\frac{4}{3} \times \frac{6}{5}}$$<br />$$n^3 = k^{\frac{8}{5}}$$<br />Now we have $n^{4a-1} = k$. Let's raise both sides to the power of $\frac{8}{5}$:<br />$$(n^{4a-1})^{\frac{8}{5}} = k^{\frac{8}{5}}$$<br />$$n^{\frac{8}{5}(4a-1)} = k^{\frac{8}{5}}$$<br />Since $n^3 = k^{\frac{8}{5}}$, we can substitute this into the equation above:<br />$$n^{\frac{8}{5}(4a-1)} = n^3$$<br />Therefore, we must have:<br />$$\frac{8}{5}(4a-1) = 3$$<br />$$8(4a-1) = 15$$<br />$$32a - 8 = 15$$<br />$$32a = 23$$<br />$$a = \frac{23}{32}$$<br /><br />Final Answer: The final answer is $\boxed{23/32}$