Question

If \(\log(x^2y^4) = a\) and \(\log \frac{x}{y} = b\), find the value of \(\frac{\log x}{\log y}\).

• A. \(\dfrac{a+4b}{a-2b}\)
• B. \(\dfrac{a-4b}{a+2b}\)
• C. \(\dfrac{a+4b}{a+2b}\)
• D. \(\dfrac{a-4b}{a-2b}\)

Hint:

Let \(p=\log x, q=\log y\); solve \(2p+4q=a\) and \(p-q=b\) together.

Answer 1

The Correct Option is A

Step 1: Rewrite the given equations using log rules.
Let \(p=\log x\) and \(q=\log y\) to keep things simple.
Using \(\log(x^2y^4) = \log x^2 + \log y^4 = 2\log x + 4\log y\), the first equation becomes
\[ 2p + 4q = a \]
Using \(\log\frac{x}{y} = \log x - \log y\), the second equation becomes
\[ p - q = b \]

Step 2: Solve the second equation for \(p\).
\[ p = b + q \]

Step 3: Substitute into the first equation and solve for \(q\).
\[ 2(b+q) + 4q = a \]
\[ 2b + 2q + 4q = a \]
\[ 6q = a - 2b \]
\[ q = \frac{a-2b}{6} \]

Step 4: Find \(p\) using the value of \(q\).
\[ p = b + q = b + \frac{a-2b}{6} = \frac{6b + a - 2b}{6} = \frac{a+4b}{6} \]

Step 5: Form the required ratio.
\[ \frac{\log x}{\log y} = \frac{p}{q} = \frac{\frac{a+4b}{6}}{\frac{a-2b}{6}} = \frac{a+4b}{a-2b} \]
The two \(6\)'s in the numerator and denominator cancel out directly.

Step 6: Why the other options are wrong.
Options (b) and (d) flip the sign in front of \(4b\), which would happen if the equation \(p-q=b\) were substituted with the wrong sign. Option (c) has the correct numerator but the wrong denominator sign, mixing up \(a+2b\) with \(a-2b\) from an arithmetic slip in Step 3.

Final Answer:
\(\dfrac{\log x}{\log y} = \dfrac{a+4b}{a-2b}\), option (a). \[ \boxed{\dfrac{a+4b}{a-2b}} \]