document.write( "Question 1206690: If:
\n" ); document.write( "log₂x + log₄x + log₁₆x = 21/4
\n" ); document.write( "find x. \n" ); document.write( "

Algebra.Com's Answer #844275 by greenestamps(13327) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "Ouch!!!

\n" ); document.write( "The solution method shown by the other tutor is valid, but WAY more work than needed....

\n" ); document.write( " $\"log%282%2C%28x%29%29%2Blog%284%2C%28x%29%29%2Blog%2816%2C%28x%29%29=21%2F4\"$

\n" ); document.write( "The bases of the logarithms are all powers of 2, making for a quick and easy path to the answer.

\n" ); document.write( " $\"16=2%5E4\"$ , so $\"log%282%2C%28x%29%29=4%2Alog%2816%2C%28x%29%29\"$
\n" ); document.write( " $\"16=4%5E2\"$ , so $\"log%284%2C%28x%29%29=2%2Alog%2816%2C%28x%29%29\"$

\n" ); document.write( "So

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\n" ); document.write( "Then

\n" ); document.write( " $\"7%2Alog%2816%2C%28x%29%29=21%2F4\"$
\n" ); document.write( " $\"log%2816%2C%28x%29%29=3%2F4\"$
\n" ); document.write( " $\"x=16%5E%283%2F4%29=8\"$

\n" ); document.write( "ANSWER: 8

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