|log{xx_{1}}| + |log{xx_{2}}| + …+ |log{xx_{n}}| + |log{x/x_{1}}| + |log{x/x_{2}}| + …+|log{x/x_{n}}|= |log{x_{1}}+ log{x_{2}}+ ….+log{x_{n}}|

Solve the following :

Find all positive real numbers x, x_{1}, x_{2}, \ldots, x_{n} such that

|\log{xx_{1}}|+|\log{xx_{2}}| + \ldots + |\log{xx_{n}}| + |\log{\frac{x}{x_{1}}}| + |\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}|= |\log{x_{1}}+ \log{x_{2}}+\log{x_{3}}+ \ldots + \log{x_{n}}|
…let us say this is given equality A

Solution:

Use the following inequality: |a-b| \leq |a| + |b| with equality iff ab \leq 0

So, we observe that : |\log{xx_{1}}|+|\log{\frac{x}{x_{1}}}| \geq |\log{xx_{1}}-\log{\frac{x}{x_{1}}}| = |\log{x_{1}^{2}}|=2 |\log{x_{1}}|,

Hence, LHS of the given equality is greater than or equal to:

2(|\log{x_{1}}|+|\log{x_{2}}|+|\log{x_{3}}|+ \ldots + |\log{x_{n}}|)

Now, let us consider the RHS of the given equality A:

we have to use the following standard result:

|\pm a_{} \pm a_{2} \pm a_{3} \ldots \pm a_{n}| \leq |a_{1}|+|a_{2}| + \ldots + |a_{n}|

So, applying the above to RHS of A:

|\log{x_{1}}+\log{x_{2}}+\ldots + \log{x_{n}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|.

But, RHS is equal to LHS as given in A:

That is, |\log{xx_{1}}|+|\log{xx_{2}}|+ \ldots + |\log{xx_{n}}| +|\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}|+ \ldots + |\log{\frac{x}{x_{n}}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|

Now, just a few steps before we proved that LHS is also greater than or equal to : That is,

|\log{xx_{1}}|+|\log{xx_{2}}|+\ldots + |\log{xx_{n}}|+ |\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}| \geq 2(|\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|)

The above two inequalities are like the following: x \leq y and x \geq 2y; so what is the conclusion? The first inequality means x2y or x=2y; clearly it means the only valid solution is x=2y.

Using the above brief result, we have here:

|\log{x_{1}}|+|\log{x_{2}}|+ \ldots +|\log{x_{n}}| =2(|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|)

Hence, we get |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|=0, which in turn means that (by applying the definition of absolute value):

|\log{x_{1}}|=|\log{x_{2}}|= \ldots =|\log{x_{n}}|, which implies that x_{1}=x_{2}= \ldots  x_{n}=1.

Substituting these values in the given logarithmic absolute value equation, we get:

n \times |\log{x}|+ n \times |\log{x}|=0, that is 2n \times |\log{x}|=0, and as n \neq 0, this implies that |\log{x}|=0 which in turn means x=1 also.

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