21 Given that \( M=\frac{18^{4 n} \times 2^{3\left(n^{2}-6 n\right)} \times 3^{2(1-4 n)}}{12^{2}} \) find the values of \( n \) for which \( M=2 \)

To find the values of \( n \) for which \( M=2 \), we need to solve the equation \( M=2 \) for \( n \). Given: \[ M=\frac{18^{4 n} \times 2^{3\left(n^{2}-6 n\right)} \times 3^{2(1-4 n)}}{12^{2}} \] Substitute \( M=2 \) into the equation: \[ 2=\frac{18^{4 n} \times 2^{3\left(n^{2}-6 n\right)} \times 3^{2(1-4 n)}}{12^{2}} \] Now, we can solve this equation for \( n \). Solve the equation by following steps: - step0: Solve for \(n\): \(2=\frac{18^{4n}\times 2^{3\left(n^{2}-6n\right)}\times 3^{2\left(1-4n\right)}}{12^{2}}\) - step1: Simplify: \(2=2^{-14n+3n^{2}-4}\) - step2: Swap the sides: \(2^{-14n+3n^{2}-4}=2\) - step3: Set the exponents equal: \(-14n+3n^{2}-4=1\) - step4: Move the expression to the left side: \(-14n+3n^{2}-4-1=0\) - step5: Subtract the numbers: \(-14n+3n^{2}-5=0\) - step6: Factor the expression: \(\left(n-5\right)\left(3n+1\right)=0\) - step7: Separate into possible cases: \(\begin{align}&n-5=0\\&3n+1=0\end{align}\) - step8: Solve the equation: \(\begin{align}&n=5\\&n=-\frac{1}{3}\end{align}\) - step9: Rewrite: \(n_{1}=-\frac{1}{3},n_{2}=5\) The values of \( n \) for which \( M=2 \) are \( n=-\frac{1}{3} \) and \( n=5 \).