Q. 53 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{l}\ln x=4 \ln y\\ {\log }_{3}x=2+2{\log }_{3}y\end{array}\right.$

Step-by-Step Solution

Answer

The solution for $\left\{\begin{array}{l}\ln x=4 \ln y\\ {\log }_{3}x=2+2{\log }_{3}y\end{array}\right.$ is, $\left(81,3\right)$.

Step 1 Given that,

$\left\{\begin{array}{l}\ln x=4 \ln y\\ {\log }_{3}x=2+2{\log }_{3}y\end{array}\right.$

Consider the  first equation,

$\ln x=4\ln y$

we can rewrite it as,

$$\begin{gathered} \ln x=\ln y^4 \\ x=y^4 \end{gathered}$$

We know that,${\log }_{3}9=2$.

Substitute the known value in the second equation.

${\log }_{3}x={\log }_{3}9+2{\log }_{3}y$.

Step 2 Now simplify it.

$$\begin{gathered} {\log }_{3}x-{\log }_{3}9+2{\log }_{3}y=0 \\ {\log }_{3}x-{\log }_{3}9+{\log }_3{y}^2=0 \end{gathered}$$

Now combine the second and third term of the equation and also replace $0$ with ${\log }_{3}1$.

$$\begin{gathered} {\log }_{3}x-{\log }_{3}9y^2={\log }_{3}1 \\ {\log }_3\left(\frac{x}{9y^2}\right)={\log }_{3}1 \\ \frac{x}{9y^2}=1 \\ x=9y^2 \end{gathered}$$

Step 3 Now equate x = y 4 and x = 9 y 2 .

$$\begin{gathered} y^4=9y^2 \\ \frac{y^4}{y^2}=\frac{9y^2}{y^2} \\ {y}^2=9 \\ y=\pm 3 \end{gathered}$$

Since the logarithm of a negative value cannot be defined, $y\ne -3$.

So, $y=3$.

Step 4 We know that, x = y 4 .

So, $x={\left(3\right)}^4$

$x=81$

The solution for the given system of equations is, $\left(81,3\right)$.