- #1

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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.

- Thread starter Wiz14
- Start date

- #1

- 20

- 0

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.

- #2

- 606

- 1

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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