Find the product $xy$ if $\newline$ $\begin{cases} 2^{x}+3^{y}=5 \ 2^{x+2}+3^{y+1}=18 \end{cases}$

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Multiply two decimals: where does the decimal point go?

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Q. Find the product

xy

\newline

\begin{cases} 2^{x}+3^{y}=5 \ 2^{x+2}+3^{y+1}=18 \end{cases}

Analyze Equation: Analyze the first equation $2^{x} + 3^{y} = 5$ . We need to find values of $x$ and $y$ that satisfy this equation. Since $5$ is a small number, we can guess and check small integer values for $x$ and $y$ .
Test Small Values: Test small integer values for $x$ and $y$ in the first equation. $\newline$ If $x = 1$ , then $2^{1} = 2$ , and we need $3^{y}$ to be $3$ to make the sum $5$ . This means $y = 1$ . $\newline$ So, one possible solution is $x = 1$ and $y = 1$ .
Verify First Solution: Verify the solution $x = 1$ and $y = 1$ in the second equation $2^{(x+2)} + 3^{(y+1)} = 18$ . Substitute $x = 1$ into $2^{(x+2)}$ to get $2^{(1+2)} = 2^3 = 8$ . Substitute $y = 1$ into $3^{(y+1)}$ to get $3^{(1+1)} = 3^2 = 9$ . Now, add these two results to see if they equal $18$ : $y = 1$ $0$ . This does not satisfy the second equation, so $x = 1$ and $y = 1$ is not the correct solution.
Try Other Values: Since $x = 1$ and $y = 1$ is not a solution, we need to try other small integer values. $\newline$ If $x = 0$ , then $2^{0} = 1$ , and we need $3^{y}$ to be $4$ to make the sum $5$ . However, $3^{y}$ cannot be $4$ for any integer $y$ , so $y = 1$ $0$ cannot be $y = 1$ $1$ . $\newline$ If $y = 1$ $2$ , then $y = 1$ $3$ , and we need $3^{y}$ to be $y = 1$ $5$ to make the sum $5$ . This means $y = 1$ $7$ . $\newline$ So, another possible solution is $y = 1$ $2$ and $y = 1$ $7$ .
Verify Second Solution: Verify the solution $x = 2$ and $y = 0$ in the second equation $2^{(x+2)} + 3^{(y+1)} = 18$ . Substitute $x = 2$ into $2^{(x+2)}$ to get $2^{(2+2)} = 2^4 = 16$ . Substitute $y = 0$ into $3^{(y+1)}$ to get $3^{(0+1)} = 3^1 = 3$ . Now, add these two results to see if they equal $18$ : $y = 0$ $0$ . This does not satisfy the second equation, so $x = 2$ and $y = 0$ is not the correct solution.
Final Attempt: Since $x = 2$ and $y = 0$ is not a solution, we need to try other small integer values. $\newline$ If $x = -1$ , then $2^{-1} = \frac{1}{2}$ , which is not an integer and will not help us find an integer solution for $y$ . Therefore, $x$ cannot be $-1$ . $\newline$ If $x = 2$ , then $2^{2} = 4$ , and we need $3^{y}$ to be $y = 0$ $0$ to make the sum $y = 0$ $1$ . This means $y = 0$ . $\newline$ We have already tried this combination and it did not work, so we made a mistake by considering it again.