document.write( "Question 630651: What's the solution for log64 1/4=x? \n" ); document.write( "

Algebra.Com's Answer #397095 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"log%2864%2C+%281%2F4%29%29+=+x\"$
\n" ); document.write( "The key here is to notice that both 64 and 1/4 are powers of 4: $\"64+=+4%5E3\"$ and $\"1%2F4+=+4%5E%28%28-1%29%29\"$ . So this expression will simplify quickly if we change the base to 4.

\n" ); document.write( "The change of base formula is
\n" ); document.write( " $\"log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29\"$
\n" ); document.write( "Using this on our expression, to change the base from 64 to 4, we get:
\n" ); document.write( " $\"log%284%2C+%281%2F4%29%29%2Flog%284%2C+%2864%29%29+=+x\"$
\n" ); document.write( "Since we know what powers of 4 64 and 1/4 are, we can substitute in for those logs:
\n" ); document.write( " $\"%28-1%29%2F3+=+x\"$
\n" ); document.write( "And we're finished! \n" ); document.write( "

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