document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=958412>Question 958412</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Garmin's yearly net sales between 2000 and 2008 can be modeled by y=191,525e^0.3162x thousand dollars (sic), where \"x\" is the number of years since 2000.\r<BR>\n" );
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document.write( "A. Write the logarithmic form of this equation.<BR>\n" );
document.write( "B. Use the logarithmic form to find when Garmin's yearly net sales will reach $3,600,000,000.\r<BR>\n" );
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document.write( "I attempted to do a few different things here, based on lecture notes and my extremely meager understanding of this level of algebra; first, I decided to give myself a base-line to work from. I know that the standard form of the equation should tell me how many years it will take to get to $3.6 billion. I'm slightly confused at the wording of the question, which states \"191,525e^0.3162x thousand dollars\" which makes me wonder - are they truncating \"millions\" of dollars into \"thousands\" to make the numbers more manageable? This seems to be case, since a Google search reveals that Garmin made more than $130 million in 2000, so an equation saying they made $191 million sounds plausible.\r<BR>\n" );
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document.write( "So, assuming that this oddly-worded sentence literally wants me to add \"thousand\" to the end of my numbers, then I can verify that Garmin made $191 million in 2000 by substituting 0 in for \"x\" (0 representing 2000): \r<BR>\n" );
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document.write( "y=191,525e^0.3162(0)\r<BR>\n" );
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document.write( "e^0.3162(0)=1 so the equation works out to y=191,525 meaning that in 2000, Garmin made $191,525 thousand dollars, or $191,525,000. So, equation verified. \r<BR>\n" );
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document.write( "So to find out how many months it takes to reach $3.6 billion with the standard equation, plug in $3.6 billion (in \"thousands\") in for y:\r<BR>\n" );
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document.write( "3,600,000=191,525e^0.3162x  and solve for \"x\". This evaluates to x=9.27789619 so I know that Garmin will net $3.6 billion sometime in early 2009.\r<BR>\n" );
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document.write( "Here's where things get confusing for me. When I write the standard equation in logarithmic form based on what I remember from lecture, I get something like: \r<BR>\n" );
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document.write( "ln(191,525)+0.3162x\r<BR>\n" );
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document.write( "which is pretty clearly not correct, since by plugging in numbers and evaluating, I get something that isn't even close to 9 years.\r<BR>\n" );
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document.write( "So obviously I'm missing an important step. An online calculator tells me that the correct format of the logarithm is ln(y)=(0.3162x). \r<BR>\n" );
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document.write( "However, plugging in the numbers for the variables, it doesn't even seem like the log format that the calculator gives me is correct:\r<BR>\n" );
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document.write( "ln(y)=(0.3162x) gives me 1 if I plug in 0 for \"x\" when it should be giving me 191,525, and plugging in 3,600,000 for \"y\" gets me 47.74334 which isn't even close to 9 years.\r<BR>\n" );
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document.write( "I am COMPLETELY lost at this point. I clearly don't understand the correct way to put this equation into logarithmic form, OR how to use the logarithmic form of the equation to solve a simple problem. I would greatly appreciate any help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #585759 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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/cgi-bin/show-face.mpl?user=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=585759\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The definition of logarithm is<BR>\n" );
document.write( "given x = b^y then y = log base b of x<BR>\n" );
document.write( "we are given <BR>\n" );
document.write( "y=191,525e^0.3162x<BR>\n" );
document.write( "note that y and x are interchanged in this example, therefore<BR>\n" );
document.write( "0.3162x = ln (y/191,525)<BR>\n" );
document.write( "use logarithm rule for division<BR>\n" );
document.write( "0.3162x = ln(y) - ln(191,525)<BR>\n" );
document.write( "calculate the natural log (ln) of 191,525 <BR>\n" );
document.write( "A) 0.3162x = ln(y) - 12.162773627<BR>\n" );
document.write( "x = (ln(y) - 12.162773627) / 0.3162<BR>\n" );
document.write( "now continue calculations, I broke them out so as to clarify process<BR>\n" );
document.write( "0.3162x = ln(3,600,000,000) - 12.162773627<BR>\n" );
document.write( "calculate the natural log (ln) of 3,600,000,000<BR>\n" );
document.write( "0.3162x = 22.004199682 - 12.162773627<BR>\n" );
document.write( "0.3162x = 9.841426055<BR>\n" );
document.write( "B) x = 31.12405457 approx 31 <BR>\n" );
document.write( "x represents number of years since 2000, therefore<BR>\n" );
document.write( "The year sales reach $3,600,000,000 is 2031<BR>\n" );
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