document.write( "Question 238165: log5(log5 x)=1 \n" ); document.write( "

Algebra.Com's Answer #175150 by jsmallt9(3758) $\"\"$ $\"About$

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When the variable you're solving for is in the argument of a logarithm, the most common way to solve the problem is to rewrite the logarithmic equation in exponential form. With this problem's log of a log we will have to do thsi twice.

\n" ); document.write( "Remember that $\"log%28a%2C+%28b%29%29+=+x\"$ is equivalent to $\"b+=+a%5Ex\"$ . We will use this twice in succession on your equation:
\n" ); document.write( " $\"log%285%2C+%28log%285%2C+%28x%29%29%29%29+=+1\"$
\n" ); document.write( "which in exponential form is:
\n" ); document.write( " $\"log%285%2C+%28x%29%29+=+5%5E1\"$
\n" ); document.write( "which simplifies to
\n" ); document.write( " $\"log%285%2C+%28x%29%29+=+5\"$
\n" ); document.write( "Rewriting this in exponential form we get:
\n" ); document.write( " $\"x+=+5%5E5\"$
\n" ); document.write( "which simplifies to
\n" ); document.write( " $\"x+=+3125\"$ \r
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