Question

A possible solution of the system of equations \[ x^2 - 8xy + 16y^2 = 0 \] \[ (\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2 = 4 \] is

• A. $x = \frac{1}{5}, y = \frac{1}{20}$
• B. $x = 100, y = 25$
• C. $x = 40, y = 10$
• D. $x = \frac{4}{25}, y = \frac{1}{16}$

Hint:

Whenever you see quadratic forms, look for perfect square factorizations first.
Always verify the domain constraints of logarithmic terms, ensuring both variables remain positive.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The given problem asks us to find a possible solution to a system of two non-linear equations.
The first equation is a quadratic form in two variables, while the second is a quadratic expression involving common logarithms.
We will solve this system by simplifying both equations into linear or simpler algebraic relations.

Step 2: Key Formula or Approach:
We will use the perfect square factorization formula:
\[ a^2 - 2ab + b^2 = (a - b)^2 \]
And the logarithmic product property:
\[ \log_{10} x + \log_{10} y = \log_{10}(xy) \]

Step 3: Detailed Explanation:

• First, let us analyze the first equation:
\[ x^2 - 8xy + 16y^2 = 0 \]
We can rewrite this expression as a perfect square:
\[ (x - 4y)^2 = 0 \implies x = 4y \]
This gives us a direct linear relationship between $x$ and $y$.

• Next, let us analyze the second equation:
\[ (\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2 = 4 \]
This is also a perfect square in terms of logarithmic values:
\[ (\log_{10} x + \log_{10} y)^2 = 4 \]

• Taking the square root on both sides:
\[ \log_{10} x + \log_{10} y = \pm 2 \]
Using the property of logarithms, we can combine the terms:
\[ \log_{10}(xy) = \pm 2 \]
This gives us two cases:
Case 1: $xy = 10^2 = 100$.
Case 2: $xy = 10^{-2} = \frac{1}{100}$.

• Let us substitute $x = 4y$ into Case 1:
\[ (4y)y = 100 \implies 4y^2 = 100 \implies y^2 = 25 \implies y = 5 \]
(Note: we only consider positive values because the domain of logarithmic functions requires $x \gt 0$ and $y \gt 0$).
Since $y = 5$, we have $x = 4(5) = 20$. So, $(20, 5)$ is a solution.

• Let us substitute $x = 4y$ into Case 2:
\[ (4y)y = \frac{1}{100} \implies 4y^2 = \frac{1}{100} \implies y^2 = \frac{1}{400} \implies y = \frac{1}{20} \]
Since $y = \frac{1}{20}$, we have $x = 4\left(\frac{1}{20}\right) = \frac{1}{5}$. So, $\left(\frac{1}{5}, \frac{1}{20}\right)$ is another solution.



Step 4: Final Answer:
Comparing our solutions with the given options, we find that option (A) matches the solution $\left(\frac{1}{5}, \frac{1}{20}\right)$.