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Solving Log Equation

Date: 04/20/2008 at 11:43:11
From: Jen
Subject: Finding log(log(x))

Is there a rule for finding log(log(x)), or for finding 
log(x + log(x))?

I have gotten to this equation:

log(x) + log(x+log(x)) + log(x+log(x) + log[x+log(x)]) = 0

I need to be able to solve for x.

Date: 04/21/2008 at 14:38:19
From: Doctor Ali
Subject: Re: Finding log(log(x))

Hi Jen,

Thanks for writing to Dr. Math.

There are no general formulas for making log(log(x)) or 
log(x+log(x)) simpler.  However, your equation is interesting.

In your equation, the sum of three logarithms is zero.  This means
that if all of them are not zero, some are positive and some are negative.

Where is log(x) positive and where is it negative?  (Just remember
that by logs, we mean logarithms in base 10.)

  For x > 1, log(x) > 0
  For x < 1, log(x) < 0
  For x = 1, log(x) = 0

First, let's assume that x > 1 and analyze the three parts of your

*) 1st LOG:

The first thing we can deduce is that:

  log(x) > 0

**) 2nd LOG:

We know that x > 1 and also log(x) > 0.  What can we conclude?  That 

  x + log(x) > 1 

since when you add a positive value to something more than one, the
result is still more than one.


  x + log(x) > 1 and log(x + log(x)) > 0

***) 3rd LOG:

Let A be x + log(x).

The third log will then be log(A + log(A))

We know that A > 1. So,

  log(A) > 0

If we add the two inequalities together, we get that:

  A + log(A) > 1

and we can deduce that the third log in the equation is also positive.

In summary, when we assume that x > 1, all of the three logs are 
positive.  We know that the sum of three positive values can never be
zero.  Thus, there are no roots for this equation if we assume that 
x > 1.

You can try a similar approach for x < 1.  After considering x > 1 and
x < 1, the only value that remains is x = 1.  For x = 1, you can just
plug x = 1 in the equation and see if it works.

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum 

Date: 04/21/2008 at 16:34:02
From: Jen
Subject: Thank you (Finding log(log(x)))

Thank you very much for the help.  I actually figured this out this
morning, but it was good to see that my logic was not faulty.
Associated Topics:
High School Logs

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