Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 852: 80


The simplified solution set for the system of equations is $\left\{ \left( {{10}^{4}},5 \right) \right\}$.

Work Step by Step

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}}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.