Question

If $x={{\log }_{2a}}a,y={{\log }_{3a}}2a\text{ and z=lo}{{\text{g}}_{4a}}3a,\text{ then }xyz+1=$
a)2yz
b)2xy
c)2zx
d)None of these

Answer 1

Hint: For solving this problem, we first consider the given part in the problem statement to simplify it. After obtaining a simplified result, we try to transform the results into given options. Using this methodology, we can easily solve the problem.

Complete step-by-step answer:

According to the problem statement, we are given $x={{\log }_{2a}}a,y={{\log }_{3a}}2a\text{ and z=lo}{{\text{g}}_{4a}}3a$.

Now, to evaluate xyz + 1, we put the values of x, y and z from the given part in the expression to obtain equation (1).

${{\log }_{2a}}a\times {{\log }_{3a}}2a\times \text{lo}{{\text{g}}_{4a}}3a+1\ldots (1)$

One of the properties of logarithmic function can be stated as: ${{\log }_{m}}n=\dfrac{{{\log

}_{e}}n}{{{\log }_{e}}m}\text{ or }\dfrac{\log n}{\log m}$

Now, by using this property we expand equation (1) to get:

$\dfrac{\log a}{\log 2a}\times \dfrac{\log 2a}{\log 3a}\times \dfrac{\log 3a}{\log 4a}+1$

Now, cancelling all the common terms from numerator and denominator, we get

$\begin{align}

  & \dfrac{\log a}{\log 4a}+1 \\

 & \dfrac{\log a+\log 4a}{\log 4a}\ldots (2) \\

\end{align}$

Another property of logarithmic function can be stated as: ${{\log }_{e}}m+{{\log }_{e}}n={{\log }_{e}}mn\text{ or }\log mn$

Operating the above-mentioned property in equation (2), we get

$\begin{align}

  & \dfrac{\log a+\log 4a}{\log 4a}=\dfrac{\log a\times 4a}{\log 4a} \\

 & \Rightarrow \dfrac{\log 4{{a}^{2}}}{\log 4a}\ldots (3) \\

\end{align}$

Another property of logarithmic function can be stated as: $\log {{m}^{n}}=n\log m$

Operating the above-mentioned property in equation (3), we get

$\begin{align}

  & \dfrac{\log 4{{a}^{2}}}{\log 4a}=\dfrac{\log {{\left( 2a \right)}^{2}}}{\log 4a} \\

 & \Rightarrow \dfrac{2\log 2a}{\log 4a}\ldots (4) \\

\end{align}$

We can manipulate the above equation (4) by multiplying and dividing the numerator by log

3a.

$\begin{align}

  & 2\times \dfrac{\log 2a}{\log 4a}=2\times \dfrac{\log 2a}{\log 3a}\times \dfrac{\log

3a}{\log 4a} \\

 & 2{{\log }_{3a}}2a\times {{\log }_{4a}}3a\ldots (5) \\

\end{align}$

Now, replacing the values of equation (5) with y and z, we get

$2{{\log }_{3a}}2a\times {{\log }_{4a}}3a=2yz$

So, finally we evaluated xyz + 1 as 2yz.

Hence, option (a) is correct.

Note: The key concept involved in solving this problem is the knowledge of properties of logarithmic functions. By applying suitable properties and manipulating the result, we obtained our answer without any error.