To solve the given system of equations:
[tex]\[
\begin{array}{l}
e^{5x - y} = 3e^{3x} \\
e^{2x} = 5e^{x + y}
\end{array}
\][/tex]
we need to find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. Let's approach this step-by-step:
1. Equation 1:
[tex]\[
e^{5x - y} = 3e^{3x}
\][/tex]
Equation 2:
[tex]\[
e^{2x} = 5e^{x + y}
\][/tex]
2. We can start by taking the natural logarithm on both sides of each equation to simplify them.
For Equation 1:
[tex]\[
\ln(e^{5x - y}) = \ln(3e^{3x})
\][/tex]
Using the properties of logarithms, [tex]\(\ln(ab) = \ln(a) + \ln(b)\)[/tex] and [tex]\(\ln(e^a) = a\)[/tex], we get:
[tex]\[
5x - y = \ln(3) + 3x
\][/tex]
Simplifying, we obtain:
[tex]\[
5x - y = \ln(3) + 3x \\
5x - 3x - y = \ln(3) \\
2x - y = \ln(3) \quad \text{(Equation 3)}
\][/tex]
3. For Equation 2:
[tex]\[
\ln(e^{2x}) = \ln(5e^{x+y})
\][/tex]
Again, using logarithm properties:
[tex]\[
2x = \ln(5) + \ln(e^{x+y}) \\
2x = \ln(5) + (x + y)
\][/tex]
Simplifying, we obtain:
[tex]\[
2x - x - y = \ln(5) \\
x - y = \ln(5) \quad \text{(Equation 4)}
\][/tex]
4. Now we have simplified the system to two linear equations:
[tex]\[
\begin{array}{l}
2x - y = \ln(3) \\
x - y = \ln(5)
\end{array}
\][/tex]
5. Subtract Equation 4 from Equation 3 to eliminate [tex]\(y\)[/tex]:
[tex]\[
(2x - y) - (x - y) = \ln(3) - \ln(5) \\
2x - y - x + y = \ln(3) - \ln(5) \\
x = \ln(3) - \ln(5)
\][/tex]
Hence, we have:
[tex]\[
x = \ln\left(\frac{3}{5}\right)
\][/tex]
6. Substitute [tex]\(x = \ln\left(\frac{3}{5}\right)\)[/tex] back into Equation 4:
[tex]\[
\ln\left(\frac{3}{5}\right) - y = \ln(5)
\][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[
y = \ln\left(\frac{3}{5}\right) - \ln(5)
\][/tex]
We can use the property [tex]\(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)[/tex]:
[tex]\[
\ln\left(\frac{3}{5}\right) - \ln(5) = \ln\left(\frac{3}{5}\right) - \ln(5) \\
y = \ln(3) - \ln(5) - \ln(5) \\
y = \ln(3) - 2\ln(5)
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
x = \ln(3) - \ln(5) \quad \text{and} \quad y = \ln(3) - 2\ln(5)
\][/tex]
