Given that \( (\sqrt[3]{4})^{k}=a \), find the values of the following in terms of \( a \). \( \begin{array}{lll}\text { (a) } 4^{k} & \text { (b) } 16^{\frac{k}{3}} & \text { (c) } 2^{k+1}\end{array} \)

Given that \( (\sqrt[3]{4})^{k} = a \), we can express \( \sqrt[3]{4} \) in terms of powers of 2: \[ \sqrt[3]{4} = 4^{1/3} = (2^2)^{1/3} = 2^{2/3} \] Thus, we can rewrite the equation as: \[ (2^{2/3})^{k} = a \] Using the property of exponents, we have: \[ 2^{\frac{2k}{3}} = a \] Now, we will find the values of \( 4^{k} \), \( 16^{\frac{k}{3}} \), and \( 2^{k+1} \) in terms of \( a \). ### (a) Finding \( 4^{k} \) We know that: \[ 4 = 2^2 \] Thus, \[ 4^{k} = (2^2)^{k} = 2^{2k} \] Now, we can express \( 2^{2k} \) in terms of \( a \): Since \( 2^{\frac{2k}{3}} = a \), we can cube both sides to find \( 2^{2k} \): \[ (2^{\frac{2k}{3}})^3 = a^3 \implies 2^{2k} = a^3 \] So, \[ 4^{k} = a^3 \] ### (b) Finding \( 16^{\frac{k}{3}} \) We know that: \[ 16 = 2^4 \] Thus, \[ 16^{\frac{k}{3}} = (2^4)^{\frac{k}{3}} = 2^{\frac{4k}{3}} \] Now, we can express \( 2^{\frac{4k}{3}} \) in terms of \( a \): We already have \( 2^{\frac{2k}{3}} = a \). Therefore, we can square both sides to find \( 2^{\frac{4k}{3}} \): \[ (2^{\frac{2k}{3}})^2 = a^2 \implies 2^{\frac{4k}{3}} = a^2 \] So, \[ 16^{\frac{k}{3}} = a^2 \] ### (c) Finding \( 2^{k+1} \) We can express \( 2^{k+1} \) as: \[ 2^{k+1} = 2^k \cdot 2^1 = 2^k \cdot 2 \] To find \( 2^k \), we can use the relationship we have with \( a \): From \( 2^{\frac{2k}{3}} = a \), we can express \( 2^k \): \[ 2^k = (2^{\frac{2k}{3}})^{\frac{3}{2}} = a^{\frac{3}{2}} \] Thus, \[ 2^{k+1} = a^{\frac{3}{2}} \cdot 2 \] So, we can write: \[ 2^{k+1} = 2a^{\frac{3}{2}} \] ### Summary of Results The values in terms of \( a \) are: \[ \begin{array}{lll} \text{(a) } 4^{k} & = a^3 \\ \text{(b) } 16^{\frac{k}{3}} & = a^2 \\ \text{(c) } 2^{k+1} & = 2a^{\frac{3}{2}} \end{array} \]