# |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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