## Precalculus (6th Edition) Blitzer

The simplified solution set for the system of equations is $\left\{ \left( {{10}^{4}},5 \right) \right\}$.
We use the substitution method to solve the provided system of equations, \left\{ \begin{align} & \log {{x}^{2}}=y+3.....\text{(I)} \\ & \log x=y-1....\text{(II)} \end{align} \right. Apply the logarithm rule ${{\log }_{m}}{{a}^{2}}=2{{\log }_{m}}a$ in equation (I), \begin{align} & \log {{x}^{2}}=y+3 \\ & 2\log x=y+3 \\ & \log x=\frac{y+3}{2} \end{align} Substitute the value of $\log x$ in equation (II), \begin{align} & \log x=y-1 \\ & \frac{y+3}{2}=y-1 \\ \end{align} The above equation is in a single variable. To solve this equation, simplify as shown below: \begin{align} & \frac{y+3}{2}=y-1 \\ & y+3=2\left( y-1 \right) \\ & y+3=2y-2 \\ & y=5 \end{align} The solution to this equation is $y=5$. Substitute the values of y in equation (II) to find values for x, For, $y=5$ \begin{align} & \log x=y-1 \\ & \log x=5-1 \\ & \log x=4 \\ \end{align} To solve this logarithmic equation, transform the equation to a power of 10 on both sides, \begin{align} & \log x=4 \\ & {{10}^{\log x}}={{10}^{4}} \end{align} Apply the logarithm rule ${{10}^{{{\log }_{{}}}a}}=a$ to simplify, \begin{align} & {{10}^{\log x}}={{10}^{4}} \\ & x={{10}^{4}} \end{align} Hence, for $y=5$, we have $x={{10}^{4}}$.