Chapter 11.6, Problem 53AYU

To determine

To solve: The given system of equation.

Answer to Problem 53AYU

Solution:

$\left(81, 3\right)$.

Explanation of Solution

Given:

$\ln x = 4 \ln y$   -----(Eq 1)

$\text{lo}{\text{g}}_{3}x = 2 + 2 {\log }_{3}y$   -----(Eq 2)

Formula used:

$\begin{array}{lll}\hfill \text{lo}{\text{g}}_{b}M& = \hfill & N\hfill \\ \hfill \Rightarrow \text{M}& = \hfill & b^N\hfill \\ \hfill \text{lo}{\text{g}}_b \left(negative number\right)& = \hfill & undefined\hfill \\ \hfill \text{lo}{\text{g}}_b\left(0\right)& = \hfill & undefined\hfill \end{array}$

Calculation:

Using above formulas Eq 1 and Eq 2 can be written as,

$x = y^4$

$\text{lo}{\text{g}}_{3}x - 2 \text{lo}{\text{g}}_{3}y = 2$

$x/y^2 = 9$

$x - 9y^2 = 0$

Substituting $x = y^4$ implies $y^4 - 9y^2 = 0$

$y^2\left(y^2 - 9\right) = 0$

Hence, $y = 3$

Substituting $x$ in $x = y^4$, $x = 81$

The solution to the equations is $\left(81, 3\right)$.