Solve {l}{a=2}{b=1/2}{c={({(a^{2}b^{3})}^{2}*{(a^3b^2)}^2)}^2}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { a = 2 }\\ { b = \frac{1}{2} }\\ { c = {({(a ^ {2} b ^ {3})} ^ {2} \cdot {(a ^ {3} b ^ {2})} ^ {2})} ^ {2} }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for a, b, c, d

$$d=1$$

Solution Steps

Consider the third equation. Insert the known values of variables into the equation.

$$c=\left(\left(2^{2}\times \left(\frac{1}{2}\right)^{3}\right)^{2}\times \left(2^{3}\times \left(\frac{1}{2}\right)^{2}\right)^{2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$c=\left(\left(4\times \left(\frac{1}{2}\right)^{3}\right)^{2}\times \left(2^{3}\times \left(\frac{1}{2}\right)^{2}\right)^{2}\right)^{2}$$

Calculate \frac{1}{2} to the power of 3 and get \frac{1}{8}.

$$c=\left(\left(4\times \frac{1}{8}\right)^{2}\times \left(2^{3}\times \left(\frac{1}{2}\right)^{2}\right)^{2}\right)^{2}$$

Multiply 4 and \frac{1}{8} to get \frac{1}{2}.

$$c=\left(\left(\frac{1}{2}\right)^{2}\times \left(2^{3}\times \left(\frac{1}{2}\right)^{2}\right)^{2}\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$c=\left(\frac{1}{4}\times \left(2^{3}\times \left(\frac{1}{2}\right)^{2}\right)^{2}\right)^{2}$$

Calculate 2 to the power of 3 and get 8.

$$c=\left(\frac{1}{4}\times \left(8\times \left(\frac{1}{2}\right)^{2}\right)^{2}\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$c=\left(\frac{1}{4}\times \left(8\times \frac{1}{4}\right)^{2}\right)^{2}$$

Multiply 8 and \frac{1}{4} to get 2.

$$c=\left(\frac{1}{4}\times 2^{2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$c=\left(\frac{1}{4}\times 4\right)^{2}$$

Multiply \frac{1}{4} and 4 to get 1.

$$c=1^{2}$$

Calculate 1 to the power of 2 and get 1.

$$c=1$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$d=1$$

The system is now solved.

$$a=2 b=\frac{1}{2} c=1 d=1$$