How to Find c in a Complex Logarithmic Equation?

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SUMMARY

The discussion centers on solving the inequality defined by the logarithmic expression log2004(log2003(log2002(log2001x))) to find the value of c, where x must be greater than c. The established solution is c = 2001^2002. The process involves ensuring that each logarithmic function is defined for positive real numbers, leading to the conclusion that log2001(x) must exceed 2002, which ultimately determines the value of c.

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Wiz14
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The Following is defined,

log2004(log2003(log2002(log2001x)))

where x > c, what is c?

Answer is 2001^2002, but how to obtain it?

I do not know much about logs as I am only in precalculus.
 
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Wiz14 said:
The Following is defined,

log2004(log2003(log2002(log2001x)))

where x > c, what is c?

Answer is 2001^2002, but how to obtain it?

I do not know much about logs as I am only in precalculus.


Remember that (the real) logarithm function at any base is defined only on the real positive numbers, thus it must be
$$\log_{2003}(\log_{2002}(\log_{2001}(x)))>0$$
(I'm assuming that those numbers you wrote in such an unclear way are the logarithms' bases), and from here
$$\log_{2002}(\log_{2001}(x))>1\Longrightarrow \log_{2001}(x)>2002\Longrightarrow...etc $$

DonAntonio
 

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