Answer :
To find the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] that satisfy the system of equations:
[tex]\[ \begin{cases} 5^{x - 2y} = 25 \\ \frac{3^{2x}}{3^y} = 3^4 \end{cases} \][/tex]
we will solve each equation step-by-step.
Step 1: Simplify the first equation [tex]\(5^{x - 2y} = 25\)[/tex]
Notice that [tex]\(25\)[/tex] can be written as a power of [tex]\(5\)[/tex]:
[tex]\[ 25 = 5^2 \][/tex]
So, the equation becomes:
[tex]\[ 5^{x - 2y} = 5^2 \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ x - 2y = 2 \][/tex]
Step 2: Simplify the second equation [tex]\(\frac{3^{2x}}{3^y} = 3^4\)[/tex]
Using the properties of exponents, specifically [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex], we can simplify the left side:
[tex]\[ 3^{2x - y} = 3^4 \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ 2x - y = 4 \][/tex]
Step 3: Solve the system of linear equations
Now we have the system of linear equations:
[tex]\[ \begin{cases} x - 2y = 2 \\ 2x - y = 4 \end{cases} \][/tex]
We can solve this system using substitution or elimination. Here, we will use substitution.
From the first equation, solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ x = 2 + 2y \][/tex]
Substitute [tex]\(x\)[/tex] into the second equation:
[tex]\[ 2(2 + 2y) - y = 4 \][/tex]
[tex]\[ 4 + 4y - y = 4 \][/tex]
[tex]\[ 4 + 3y = 4 \][/tex]
[tex]\[ 3y = 0 \][/tex]
[tex]\[ y = 0 \][/tex]
Substitute [tex]\(y = 0\)[/tex] back into the expression for [tex]\(x\)[/tex]:
[tex]\[ x = 2 + 2(0) \][/tex]
[tex]\[ x = 2 \][/tex]
So, the real solution to the system is:
[tex]\[ (x, y) = (2, 0) \][/tex]
Step 4: Identify any additional complex solutions
Using further algebraic techniques beyond elementary methods, we can identify additional complex solutions for this system. The full set of solutions includes complex numbers:
[tex]\[ (x, y) = \left(2, 0\right), \ \left(2 - \frac{4i\pi}{3\log(3)}, -\frac{2i\pi}{3\log(3)}\right), \ \left(2 + \frac{4i\pi}{3\log(3)}, \frac{2i\pi}{3\log(3)}\right) \][/tex]
Thus, the complete set of solutions is:
[tex]\[ \left(2, 0\right), \ \left(2 - \frac{4i\pi}{3\log(3)}, -\frac{2i\pi}{3\log(3)}\right), \ \left(2 + \frac{4i\pi}{3\log(3)}, \frac{2i\pi}{3\log(3)}\right) \][/tex]
[tex]\[ \begin{cases} 5^{x - 2y} = 25 \\ \frac{3^{2x}}{3^y} = 3^4 \end{cases} \][/tex]
we will solve each equation step-by-step.
Step 1: Simplify the first equation [tex]\(5^{x - 2y} = 25\)[/tex]
Notice that [tex]\(25\)[/tex] can be written as a power of [tex]\(5\)[/tex]:
[tex]\[ 25 = 5^2 \][/tex]
So, the equation becomes:
[tex]\[ 5^{x - 2y} = 5^2 \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ x - 2y = 2 \][/tex]
Step 2: Simplify the second equation [tex]\(\frac{3^{2x}}{3^y} = 3^4\)[/tex]
Using the properties of exponents, specifically [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex], we can simplify the left side:
[tex]\[ 3^{2x - y} = 3^4 \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ 2x - y = 4 \][/tex]
Step 3: Solve the system of linear equations
Now we have the system of linear equations:
[tex]\[ \begin{cases} x - 2y = 2 \\ 2x - y = 4 \end{cases} \][/tex]
We can solve this system using substitution or elimination. Here, we will use substitution.
From the first equation, solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ x = 2 + 2y \][/tex]
Substitute [tex]\(x\)[/tex] into the second equation:
[tex]\[ 2(2 + 2y) - y = 4 \][/tex]
[tex]\[ 4 + 4y - y = 4 \][/tex]
[tex]\[ 4 + 3y = 4 \][/tex]
[tex]\[ 3y = 0 \][/tex]
[tex]\[ y = 0 \][/tex]
Substitute [tex]\(y = 0\)[/tex] back into the expression for [tex]\(x\)[/tex]:
[tex]\[ x = 2 + 2(0) \][/tex]
[tex]\[ x = 2 \][/tex]
So, the real solution to the system is:
[tex]\[ (x, y) = (2, 0) \][/tex]
Step 4: Identify any additional complex solutions
Using further algebraic techniques beyond elementary methods, we can identify additional complex solutions for this system. The full set of solutions includes complex numbers:
[tex]\[ (x, y) = \left(2, 0\right), \ \left(2 - \frac{4i\pi}{3\log(3)}, -\frac{2i\pi}{3\log(3)}\right), \ \left(2 + \frac{4i\pi}{3\log(3)}, \frac{2i\pi}{3\log(3)}\right) \][/tex]
Thus, the complete set of solutions is:
[tex]\[ \left(2, 0\right), \ \left(2 - \frac{4i\pi}{3\log(3)}, -\frac{2i\pi}{3\log(3)}\right), \ \left(2 + \frac{4i\pi}{3\log(3)}, \frac{2i\pi}{3\log(3)}\right) \][/tex]