## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 852: 79

#### Answer

The simplified solution set for the system of equations is $\left\{ \left( 8,2 \right) \right\}$.

#### Work Step by Step

We use the substitution method to solve the provided system of equations: \left\{ \begin{align} & {{\log }_{y}}x=3....\text{(I)} \\ & {{\log }_{y}}\left( 4x \right)=5....\text{(II)} \end{align} \right. By using the logarithm rule ${{\log }_{m}}\left( ab \right)={{\log }_{m}}a+{{\log }_{m}}b$ in equation (II), we get \begin{align} & {{\log }_{y}}\left( 4x \right)=5 \\ & {{\log }_{y}}x+{{\log }_{y}}4=5 \\ & {{\log }_{y}}x=5-{{\log }_{y}}4 \end{align} Substitute this value of ${{\log }_{y}}x$ in equation (I), \begin{align} & {{\log }_{y}}x=3 \\ & 5-{{\log }_{y}}4=3 \\ & {{\log }_{y}}4=2 \end{align} The above equation is in a single variable. To solve this equation, transform the equation to the power of y on both sides. \begin{align} & {{\log }_{y}}4=2 \\ & {{y}^{{{\log }_{y}}4}}={{y}^{2}} \end{align} We use the logarithm rule ${{m}^{{{\log }_{m}}a}}=a$ to simplify as shown below: \begin{align} & {{y}^{{{\log }_{y}}4}}={{y}^{2}} \\ & 4={{y}^{2}} \\ & y=\pm 2 \end{align} The solution to this equation is $y=2$ or $y=-2$, but since the base of a logarithmic function cannot be a negative value, we ignore the solution $y=-2$. Putting the values of y in equation (I) to find values for x, we get, For $y=2$, \begin{align} & {{\log }_{y}}x=3 \\ & {{\log }_{2}}x=3 \\ \end{align} To solve for $x$ in this logarithmic equation, transform the equation to a power of 2 on both sides, \begin{align} & {{\log }_{2}}x=3 \\ & {{2}^{{{\log }_{2}}x}}={{2}^{3}} \end{align} Apply the logarithm rule ${{m}^{{{\log }_{m}}a}}=a$ to simplify, \begin{align} & {{2}^{{{\log }_{2}}x}}={{2}^{3}} \\ & x={{2}^{3}} \\ & x=8 \end{align} Hence, for $y=2$, we have $x=8$.

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