document.write( "Question 245805: ln[log2(lnx))]=0 what is x \n" ); document.write( "

Algebra.Com's Answer #179475 by jsmallt9(3758) $\"\"$ $\"About$

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.

\n" ); document.write( "Rewriting logarithmic equations in exponential form requires knowing that
\n" ); document.write( " $\"log%28a%2C+%28p%29%29+=+q\"$ is equivalent to $\"p+=+a%5Eq\"$
\n" ); document.write( "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:
\n" ); document.write( " $\"+ln%28log%282%2C+%28ln%28x%29%29%29%29+=+0\"$
\n" ); document.write( "In exponential form:
\n" ); document.write( " $\"+log%282%2C+%28ln%28x%29%29%29+=+e%5E0\"$ (Since the base of ln is e.)
\n" ); document.write( "Since any non-zero number, including e, to the zero power is 1 this simplifies to:
\n" ); document.write( " $\"+log%282%2C+%28ln%28x%29%29%29+=+1\"$
\n" ); document.write( "This equation in exponential form:
\n" ); document.write( " $\"ln%28x%29+=+2%5E1\"$
\n" ); document.write( "Since 2^1 = 2 this simplifies to:
\n" ); document.write( " $\"ln%28x%29+=+2\"$
\n" ); document.write( "This equation in exponential form:
\n" ); document.write( " $\"x+=+e%5E2\"$ \n" ); document.write( "

\n" );