document.write( "Question 308969: Use the Change of Base formula to evaluate \"+log%284%29+20+\". Then convert \"+log%284%29+20+\" to a logarithm in base 3. \r
\n" ); document.write( "\n" ); document.write( "A) 2.161; \"+log%283%29+15+\"
\n" ); document.write( "B) 2.996; \"+log%283%29+10.741+\"
\n" ); document.write( "C) 1.301; \"+log%283%29+15+\"
\n" ); document.write( "D) 2.161; \"+log%283%29+10.741+\"
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Algebra.Com's Answer #220969 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
Use the Change of Base formula to evaluate \"+log%284%29+20+\".
\n" ); document.write( "= log(20)/log(4) = 2.1610
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\n" ); document.write( "\n" ); document.write( "Then convert \"+log%284%29+20+\" to a logarithm in base 3.
\n" ); document.write( "2.1610 = log3(x)
\n" ); document.write( "x = 3^2.1610
\n" ); document.write( "x = 10.741
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.\r
\n" ); document.write( "\n" ); document.write( "=============================== \r
\n" ); document.write( "\n" ); document.write( "A) 2.161; \"+log%283%29+15+\"
\n" ); document.write( "B) 2.996; \"+log%283%29+10.741+\"
\n" ); document.write( "C) 1.301; \"+log%283%29+15+\"
\n" ); document.write( "D) 2.161; \"+log%283%29+10.741\"
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