document.write( "Question 959136: If log7(log3(log2(x)))=0, find the value of x^-1/2. \n" ); document.write( "
Algebra.Com's Answer #586284 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
First we will find x and then x^(-1/2). To find x we will repeatedly rewrite the equation in exponential form (to \"peel away\" the logs).

\n" ); document.write( "In general \"log%28a%2C+%28p%29%29+=+n\" is equivalent to \"a%5En+=+p\". Using this pattern on
\n" ); document.write( "\"log%287%2C+%28log%283%2C+%28log%282%2C+%28x%29%29%29%29%29%29=0\"
\n" ); document.write( "\"7%5E0+=+log%283%2C+%28log%282%2C+%28x%29%29%29%29%29%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"1+=+log%283%2C+%28log%282%2C+%28x%29%29%29%29%29%29\"
\n" ); document.write( "Repeating...
\n" ); document.write( "\"3%5E1+=+log%282%2C+%28x%29%29%29%29\"
\n" ); document.write( "\"3+=+log%282%2C+%28x%29%29%29%29\"
\n" ); document.write( "Again...
\n" ); document.write( "\"2%5E3+=+x\"
\n" ); document.write( "\"8+=+x\"

\n" ); document.write( "Now for \"x%5E%28-1%2F2%29\":
\n" ); document.write( "\"8%5E%28-1%2F2%29\"
\n" ); document.write( "\"1%2F8%5E%281%2F2%29\"
\n" ); document.write( "\"1%2Fsqrt%288%29\"
\n" ); document.write( "\"%281%2Fsqrt%288%29%29%28sqrt%282%29%2Fsqrt%282%29%29\"
\n" ); document.write( "\"sqrt%282%29%2Fsqrt%2816%29\"
\n" ); document.write( "\"sqrt%282%29%2F4\"
\n" ); document.write( "
\n" );