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