document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=251335>Question 251335</A>: <I><SPAN STYLE=\"word-break: break-all;\"> To measure the volume of loudness of a sound, the decibel scale is used. The loudness L in decibels (db) of the sound is given by: L =10 log I/I0 where I is the intensity of the sound.\r<BR>\n" );
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document.write( "Solve: The band Strange Folk performed in Burlington VT and reached sound levels of 111 db. What is the intensity of the sounds? This is my answer, but my teacher says I did it wrong. Here's what he wrote:\r<BR>\n" );
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document.write( "You have made some errors on this question.\r<BR>\n" );
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document.write( "L =10 log I/I0<BR>\n" );
document.write( "111 = 10 log (l/10)<BR>\n" );
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document.write( "should be<BR>\n" );
document.write( "L =10 log I/I0<BR>\n" );
document.write( "111 = 10 log (l/10^(-12)<BR>\n" );
document.write( "11.1=log I - log 10^ (-12)<BR>\n" );
document.write( "Now solve for I<BR>\n" );
document.write( "Thanks!<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #183212 by </B><A HREF=/tutors/aboutme.mpl?userid=jsmallt9><B>jsmallt9(3758)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jsmallt9><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=183212\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It took me a while to understand what was going on. The equation is<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=L+=+10log%28%28I%2FI%5B0%5D%29%29 ALIGN=MIDDLE  ALT=\"L+=+10log%28%28I%2FI%5B0%5D%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Those are i's in the logarithm, not 1's. And the zero in the logarithm is a subscript, not the digit of a number. I think this is where you got confused. (I know this is where I got confused when I first looked at the problem.)<br><BR>\n" );
document.write( "Since <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=I%5B0%5D+=+10%5E%28-12%29 ALIGN=MIDDLE  ALT=\"I%5B0%5D+=+10%5E%28-12%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, your teacher's solution:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=L+=10log%28%28I%2FI%5B0%5D%29%29 ALIGN=MIDDLE  ALT=\"L+=10log%28%28I%2FI%5B0%5D%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=111+=+10log%28%28I%2F10%5E%28-12%29%29%29 ALIGN=MIDDLE  ALT=\"111+=+10log%28%28I%2F10%5E%28-12%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=11.1=log%28%28I%29%29+-+log%28%2810%5E%28-12%29%29%29 ALIGN=MIDDLE  ALT=\"11.1=log%28%28I%29%29+-+log%28%2810%5E%28-12%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "is correct so far.<br><BR>\n" );
document.write( "And, as your teacher says, we still need to solve for I. Solving for I means we have to get an equation we I is by itself on one side.<br><BR>\n" );
document.write( "We can start by finding the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=log%28%28+10%5E%28-12%29%29%29 ALIGN=MIDDLE  ALT=\"log%28%28+10%5E%28-12%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >/ If we understand logarithms we will know that this must be -12. If you don't see this:<ul><li>Ask your calculator</li><li>Use the property of logarithms, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29 ALIGN=MIDDLE  ALT=\"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, to rewrite <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=log%28%28+10%5E%28-12%29%29%29 ALIGN=MIDDLE  ALT=\"log%28%28+10%5E%28-12%29%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-12%2Alog%28%28+10%29%29 ALIGN=MIDDLE  ALT=\"-12%2Alog%28%28+10%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >. And since <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=log%28%2810%29%29+=+1 ALIGN=MIDDLE  ALT=\"log%28%2810%29%29+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" > by definition this becomes -12.</li></ul><BR>\n" );
document.write( "Our equation is now:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=11.1=log%28%28I%29%29+-+%28-12%29 ALIGN=MIDDLE  ALT=\"11.1=log%28%28I%29%29+-+%28-12%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "or<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=11.1=log%28%28I%29%29+%2B+12 ALIGN=MIDDLE  ALT=\"11.1=log%28%28I%29%29+%2B+12\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Next we can subtract 12 from each side:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-0.9+=+log%28%28I%29%29 ALIGN=MIDDLE  ALT=\"-0.9+=+log%28%28I%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "If you know how to do inverse logarithms on your calculator you can find \"I\" now. If not, then we can rewrite this in exponential form:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=10%5E%28-0.9%29+=+I ALIGN=MIDDLE  ALT=\"10%5E%28-0.9%29+=+I\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Now we can find \"I\" by using our calculators to raise 10 to the -0.9 power. (I'll leave this step up to you.)<BR>\n" );
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