Question

If ${a^x} = {b^y} = {c^z} = {d^w}$ , then the values of $x[(\dfrac{1}{y}) + (\dfrac{1}{z}) + (\dfrac{1}{w})]$ is
A. ${\log _a}bca$
B. ${\log _a}bcd$
C. ${\log _b}acd$
D. ${\log _c}bad$

Answer 1

Hint: Before we proceed to solve our problem, it is important to know about the concept we are using. i.e. properties of logarithmic functions.

Formula Used:
Properties of Logarithmic Functions to be used is the Product Rule ${\log _b}(MN) = {\log _b}(M) + {\log _b}(M)$

Complete step by step Solution:
Given ${a^x} = {b^y} = {c^z} = {d^w}$
Taking log both sides, we get,
$ \Rightarrow \log {a^x} = \log {b^y}$
$ \Rightarrow x\log a = y\log b$
$ \Rightarrow \dfrac{x}{y} = \dfrac{{\log a}}{{\log b}}$
$ \Rightarrow {\log _a}b = \dfrac{x}{y}$-----(1)
Similarly,
${\log _a}c = \dfrac{x}{z}$-------(2)
${\log _a}d = \dfrac{x}{w}$------(3)
Now adding eq. 1, 2, 3 we get,
${\log _a}b + {\log _a}c + {\log _a}d = {\log _a}bcd$ by using logarithmic property
$\dfrac{x}{y} + \dfrac{x}{z} + \dfrac{x}{w} = {\log _a}bcd$
$x[(\dfrac{1}{y}) + (\dfrac{1}{z}) + (\dfrac{1}{w})] = {\log _a}bcd$

Hence, the correct option is B.

Additional Information:The log of a number x is the exponent to which another constant number, b also called the base, must be raised, to produce that number x. It is also known as the inverse of the exponent function.
A logarithmic function is defined as $f(x) = {\log _b}x$
This property indicates that the log of a product of two numbers is the sum of the logarithmic of its factors i.e., multiply two numbers having the same base, then add the exponents



Note: Take care while applying the logarithmic properties in this type of question. Without knowing about logarithmic properties, you cannot think about the approach. Also, while taking logs both sides ensure that the value should be real and positive.