Question

If $\dfrac{{xy\log xy}}{{x + y}} = \dfrac{{yz\log yz}}{{y + z}} = \dfrac{{zx\log zx}}{{z + x}}$ show that ${x^x} = {y^y} = {z^z}$ ?

Answer 1

Hint : Logarithms are defined as the ways to figure out which exponents and when we need to multiply to get the specific number. Here we will assume one variable for these equivalent terms and will apply different identities of the logarithms to get the required expression.

Complete step by step solution:
Take the given expression: $\dfrac{{xy\log xy}}{{x + y}} = \dfrac{{yz\log yz}}{{y + z}} = \dfrac{{zx\log zx}}{{z + x}}$
A part in the multiplicative in the numerator can be written as the denominator’s denominator.
$\Rightarrow \dfrac{{\log xy}}{{\dfrac{{x + y}}{{xy}}}} = \dfrac{{\log yz}}{{\dfrac{{y + z}}{{yz}}}} = \dfrac{{\log zx}}{{\dfrac{{z + x}}{{zx}}}}$
Segregate the denominator’s denominator and simplify by considering that the common factors from the numerator and the denominator cancels each other.
$\Rightarrow \dfrac{{\log xy}}{{\dfrac{1}{y} + \dfrac{1}{x}}} = \dfrac{{\log yz}}{{\dfrac{1}{z} + \dfrac{1}{y}}} = \dfrac{{\log zx}}{{\dfrac{1}{z} + \dfrac{1}{x}}}$
Since all the three terms in the above expression are equal, let us assume this equal to the term “k”
$\Rightarrow \dfrac{{\log xy}}{{\dfrac{1}{y} + \dfrac{1}{x}}} = \dfrac{{\log yz}}{{\dfrac{1}{z} + \dfrac{1}{y}}} = \dfrac{{\log zx}}{{\dfrac{1}{z} + \dfrac{1}{x}}} = k$
Take each term equal to “k”
$\Rightarrow \dfrac{{\log xy}}{{\dfrac{1}{y} + \dfrac{1}{x}}} = k$ ,
$\Rightarrow \dfrac{{\log yz}}{{\dfrac{1}{z} + \dfrac{1}{y}}} = k$ and
$\Rightarrow \dfrac{{\log zx}}{{\dfrac{1}{z} + \dfrac{1}{x}}} = k$
Perform cross multiplication, where the denominator of the one side is multiplied with the term in the numerator of the opposite side.
$\Rightarrow \log xy = k\left[ {\dfrac{1}{y} + \dfrac{1}{x}} \right]$
$\Rightarrow \log yz = k\left[ {\dfrac{1}{y} + \dfrac{1}{z}} \right]$
$\Rightarrow \log zx = k\left[ {\dfrac{1}{z} + \dfrac{1}{x}} \right]$
Apply Product rule: $\log xy = \log x + \log y$ in all the three terms
$\Rightarrow \log x + \log y = k\left[ {\dfrac{1}{y} + \dfrac{1}{x}} \right]$ …. (A)
$\Rightarrow \log y + \log z = k\left[ {\dfrac{1}{y} + \dfrac{1}{z}} \right]$ ….. (B)
$\Rightarrow \log z + \log x = k\left[ {\dfrac{1}{z} + \dfrac{1}{x}} \right]$ ….. (C)
Add all the three equations –
$\Rightarrow 2(\log x + \log y + \log z) = 2k\left[ {\dfrac{1}{y} + \dfrac{1}{x} + \dfrac{1}{z}} \right]$
Common factors from the above both sides of the equation cancel each other.
$\Rightarrow (\log x + \log y + \log z) = k\left[ {\dfrac{1}{y} + \dfrac{1}{x} + \dfrac{1}{z}} \right]$ ….. (D)
Subtract equation (A), (B) and (C) one by one, remember like terms with the same value and the opposite signs cancel each other.
$\Rightarrow \log z = k\left[ {\dfrac{1}{z}} \right]$ …. (E)
$\Rightarrow \log x = k\left[ {\dfrac{1}{x}} \right]$ …. (F)
$\Rightarrow \log y = k\left[ {\dfrac{1}{y}} \right]$ …. (G)
Cross multiply when the numerator of one side is multiplied with the denominator of the opposite side.
$\Rightarrow z\log z = k \\\ \Rightarrow y\log y = k \\\ \Rightarrow x\log x = k \;$
Now, apply Power rule: $\log {x^n} = n\log x$
$\Rightarrow \log {z^z} = k \\\ \Rightarrow \log {y^y} = k \\\ \Rightarrow \log {x^x} = k \;$
The above expressions can be written as –
$\Rightarrow \log {z^z} = \log {y^y} = \log {x^x}$
Remove log from all the terms –
$\Rightarrow {z^z} = {y^y} = {x^x}$
Hence, proved ${x^x} = {y^y} = {z^z}$

Note : The logarithm is defined as the power to which the number must be raised in order to get some other terms. Always remember the standard properties of the logarithm such as Product rule, quotient rule and the power rule. The basic and appropriate logarithm properties are most important and the solution totally depends on it, so remember and understand its application properly.