SOLUTION: ln[log2(lnx))]=0 what is x

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Question 245805: ln[log2(lnx))]=0 what is x
Found 2 solutions by nerdybill, jsmallt9:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
I believe your problem is:
ln%28log%282%2Cln%28x%29%29%29=0
log%282%2Cln%28x%29%29+=+e%5E0
log%282%2Cln%28x%29%29+=+1+
ln%28x%29+=+2%5E1+
ln%28x%29+=+2+
x+=+e%5E2+
x+=+7.389

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
To solve for variables in the argument of a logarithm we generally rewrite the logarithmic equation in exponential form. Since your equation has a logarithm of a logarithm of a logarithm, we will have to do this three times.

Rewriting logarithmic equations in exponential form requires knowing that
log%28a%2C+%28p%29%29+=+q is equivalent to p+=+a%5Eq
The "a", "p" and "q" can be any expression. We will use this to "peel away" the logarithms, one at a time, like peeling the layers of an onion:
+ln%28log%282%2C+%28ln%28x%29%29%29%29+=+0
In exponential form:
+log%282%2C+%28ln%28x%29%29%29+=+e%5E0 (Since the base of ln is e.)
Since any non-zero number, including e, to the zero power is 1 this simplifies to:
+log%282%2C+%28ln%28x%29%29%29+=+1
This equation in exponential form:
ln%28x%29+=+2%5E1
Since 2^1 = 2 this simplifies to:
ln%28x%29+=+2
This equation in exponential form:
x+=+e%5E2