Certainly! To solve the simultaneous equations given:
[tex]\[
\begin{array}{l}
\ln x = 2 \ln y \quad \text{(Equation 1)} \\
\ln y - \ln x = 1 \quad \text{(Equation 2)}
\end{array}
\][/tex]
we will handle each equation step by step.
### Step 1: Solve for [tex]\(\ln x\)[/tex] and [tex]\(\ln y\)[/tex]
#### Equation 1:
[tex]\[
\ln x = 2 \ln y
\][/tex]
Express [tex]\(\ln y\)[/tex] from this equation:
[tex]\[
\ln x = \ln y^2
\][/tex]
This implies:
[tex]\[
x = y^2
\][/tex]
#### Equation 2:
[tex]\[
\ln y - \ln x = 1
\][/tex]
Substitute [tex]\(\ln x\)[/tex] from Equation 1 into Equation 2:
[tex]\[
\ln y - 2 \ln y = 1
\][/tex]
[tex]\[
-\ln y = 1
\][/tex]
[tex]\[
\ln y = -1
\][/tex]
Solve for [tex]\(y\)[/tex]:
[tex]\[
y = e^{-1} = \frac{1}{e}
\][/tex]
### Step 2: Solve for [tex]\(x\)[/tex] using the value of [tex]\(y\)[/tex]
Using the relationship [tex]\(x = y^2\)[/tex]:
[tex]\[
x = \left( \frac{1}{e} \right)^2 = \frac{1}{e^2}
\][/tex]
### Step 3: Verify the solutions in both equations
Substitute [tex]\(x = \frac{1}{e^2}\)[/tex] and [tex]\(y = \frac{1}{e}\)[/tex] back into the original equations to verify they satisfy both:
#### Verify Equation 1:
[tex]\[
\ln x \stackrel{?}{=} 2 \ln y
\][/tex]
[tex]\[
\ln \left( \frac{1}{e^2} \right) = 2 \ln \left( \frac{1}{e} \right)
\][/tex]
[tex]\[
-2 \ln e = 2 (-\ln e)
\][/tex]
[tex]\[
-2 = 2 (-1)
\][/tex]
[tex]\[
-2 = -2
\][/tex]
This is true.
#### Verify Equation 2:
[tex]\[
\ln y - \ln x \stackrel{?}{=} 1
\][/tex]
[tex]\[
\ln \left( \frac{1}{e} \right) - \ln \left( \frac{1}{e^2} \right) = 1
\][/tex]
[tex]\[
-\ln e - (-2 \ln e) = 1
\][/tex]
[tex]\[
-1 + 2 = 1
\][/tex]
[tex]\[
1 = 1
\][/tex]
This is also true.
Therefore, the values [tex]\(x = \frac{1}{e^2}\)[/tex] and [tex]\(y = \frac{1}{e}\)[/tex] are the correct solutions for the given simultaneous equations.
