If `(log_(2)x)/(4)=(log_(2)y)/(6)=(log_(2)z)/(3k)` and `x^(3)y^(2)z=1`, then k= …..

Answer

To solve the given problem, we start with the equations provided:

1. \(\frac{\log_2 x}{4} = \frac{\log_2 y}{6} = \frac{\log_2 z}{3k}\)

Let us denote this common value as \(t\). Therefore, we can express the logarithms in terms of \(t\):

\[ \log_2 x = 4t \]
\[ \log_2 y = 6t \]
\[ \log_2 z = 3kt \]

2. We also have the equation:

\[ x^3 y^2 z = 1 \]

Taking the logarithm (base 2) of both sides, we get:

\[ \log_2 (x^3 y^2 z) = \log_2 1 \]

Since \(\log_2 1 = 0\), we have:

\[ \log_2 (x^3) + \log_2 (y^2) + \log_2 z = 0 \]

Using the properties of logarithms, we can simplify this:

\[ 3 \log_2 x + 2 \log_2 y + \log_2 z = 0 \]

Now substituting the values of \(\log_2 x\), \(\log_2 y\), and \(\log_2 z\) from our earlier expressions:

\[ 3(4t) + 2(6t) + (3kt) = 0 \]

This simplifies to:

\[ 12t + 12t + 3kt = 0 \]

Combining the terms, we have:

\[ 24t + 3kt = 0 \]

3. We can factor out \(t\) (assuming \(t \neq 0\)):

\[ t(24 + 3k) = 0 \]

Since \(t \neq 0\), we can set the factor equal to zero:

\[ 24 + 3k = 0 \]

4. Solving for \(k\):

\[ 3k = -24 \]
\[ k = -8 \]

Thus, the final answer is:

\[ \boxed{-8} \]