Q. 54 - Precalculus Enhanced with Graphing Utilities

Question

In Problems $25-54$, solve each system. Use any method you wish.

$\left\{\begin{array}{l}\ln x=5 \ln y\\ {\log }_{2}x=3+2{\log }_{2}y\end{array}\right.$

Step-by-Step Solution

Answer

The solution for $\left\{\begin{array}{l}\ln x=5 \ln y\\ {\log }_{2}x=3+2{\log }_{2}y\end{array}\right.$ is, $\left(32,2\right)$.

Step 1 Given that,

$\left\{\begin{array}{l}\ln x=5 \ln y\\ {\log }_{2}x=3+2{\log }_{2}y\end{array}\right.$

Consider the first equation and simplify it using the logarithmic properties.

$$\begin{gathered} \ln x=5 \ln y \\ \ln x-5 \ln y=0 \\ \ln x-\ln y^5=0 \\ \ln \left(\frac{x}{y^5}\right)=0 \\ \frac{x}{y^5}=e^0 \\ \frac{x}{y^5}=1 \\ x=y^5 \end{gathered}$$

Step 2 Now consider the second equation and simplify it by using the logarithmic properties.

$$\begin{gathered} {\log }_{2}x=3+2{\log }_{2}y \\ {\log }_{2}x=3+{\log }_2{y}^2 \\ {\log }_{2}x-{\log }_2{y}^2=3 \\ {\log }_2\left(\frac{x}{y^2}\right)=3 \\ \frac{x}{y^2}={\left(2\right)}^3 \\ x=8y^2 \end{gathered}$$

Step 3 Now equate x = y 5 and x = 8 y 2 .

$$\begin{gathered} y^5=8y^2 \\ \frac{y^5}{y^2}=\frac{8y^2}{y^2} \\ {y}^3=8 \\ y=2 \end{gathered}$$

Substitute $y=2$ in $x=y^5$.

$$\begin{gathered} x={\left(2\right)}^5 \\ x=32 \end{gathered}$$

The solution for the given system of equation is, $\left(32,2\right)$.