Question

If \[\dfrac{{{\log }_{2}}x}{4}=\dfrac{{{\log }_{2}}y}{6}=\dfrac{{{\log }_{2}}z}{3k}\] and \[{{x}^{3}}{{y}^{2}}z=1\] then \[k=\]
(a) -8
(b) -4
(c) 0
(d) \[{{\log }_{2}}\dfrac{1}{256}\]

Answer 1

Hint: In this problem we have to solve the given equation by using the logarithm formula we have, \[{{\log }_{a}}x=\dfrac{{{\log }_{c}}x}{{{\log }_{c}}a}\] simplifying the higher term till get the required term to solve the given equation for x. Then we get the value of k.

Complete step by step answer:
According to the question we have to find the value of k in the equation given.
Given \[{{x}^{3}}{{y}^{2}}z=1\]
\[\begin{align}
  & \log \left( {{x}^{3}}{{y}^{2}}z \right)=\log 1 \\
 & \log {{x}^{3}}+\log {{y}^{2}}+\log z=0 \\
\end{align}\]
And it is also given that,
\[\dfrac{{{\log }_{2}}x}{4}=\dfrac{{{\log }_{2}}y}{6}=\dfrac{{{\log }_{2}}z}{3k}\]
\[\dfrac{\log x}{{{\log }{2}}.4}=\dfrac{\log y}{{{\log }{2}}.6}=\dfrac{\log z}{{{\log }{2}}.3k}\left[ \because {{\log }_{b}}a=\dfrac{\log a}{\log b} \right]\]
So, if we solve this then we get,
Because we know that \[\log 2\] can be taken out as common from the denominator of all three of the terms of expression, therefore, we can write
\[\dfrac{\log x}{4}=\dfrac{\log y}{6}=\dfrac{\log z}{3k}\] - (1)
And, we know that the above equation can be further expressed as the following, therefore, we get
\[\dfrac{\log x}{4}=\dfrac{\log z}{3k}\]
We can write it as
\[\log x=\dfrac{4}{3k}\log 3\] - (2)
And \[\log y=\dfrac{6}{3k}\log 3\] -(3)
Substitute (2) and (3) in equation (1)
\[\begin{align}
  & 3.\dfrac{4}{3k}\log 3+2\times \dfrac{6}{3k}\log 3+\log 3=0 \\
 & \dfrac{4}{k}{{\log }{3}}+\dfrac{4}{k}{{\log }{3}}+{{\log }{3}}=0 \\
 & {{\log }{3}}\left[ \dfrac{4}{k}+\dfrac{4}{k}+1 \right]=0 \\
\end{align}\]
So, \[\Rightarrow \dfrac{8}{k}+1=0\Rightarrow k=-8\] option (a)
\[\Rightarrow -8={{\log }_{2}}{{\left( 256 \right)}^{-1}}\] option (d)

So, the correct answer is “Option a and d”.

Note: We know that \[{{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x\] which is a logarithmic identity which can be used in the problem like this if they provide in power form. When two logarithmic functions with the same base are equal then we can equate the number.