document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187529>Question 1187529</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello, I am currently doing work on some exponential and logarithmic problems.  I am stuck on a word problem and don't understand how to even begin.  Please help me solve the following problem:\r<BR>\n" );
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document.write( "A computer purchased for $750 loses 11% of its value every year.\r<BR>\n" );
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document.write( "The computer's value can be modeled by the function v(t)=a⋅b^t, where v is the dollar value and t the number of years since purchase.\r<BR>\n" );
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document.write( "A) In the exponential model a= ___ and b = ___\r<BR>\n" );
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document.write( "B) In how many years will the computer be worth half its original value? Round answer to 1 decimal place. The answer is ___ years\r<BR>\n" );
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document.write( "Thank you!\r<BR>\n" );
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document.write( "-Diana </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #818534 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=818534\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the computer is purchased for 750.<BR>\n" );
document.write( "it loses 11% of its value each year.<BR>\n" );
document.write( "the function  you are using is:<BR>\n" );
document.write( "v(t) = a * b^t<BR>\n" );
document.write( "a would be the original amount = 750<BR>\n" );
document.write( "b would be the rate that it is growing.<BR>\n" );
document.write( "if it shrinking, than b will be less than 1.<BR>\n" );
document.write( "you are given that it loses 11% of its value each year.<BR>\n" );
document.write( "11% is equal to .11.<BR>\n" );
document.write( "11 is the rate percent.<BR>\n" );
document.write( ".11 is the rate.<BR>\n" );
document.write( "rate percent / 100 = rate<BR>\n" );
document.write( "rate * 100 = rate percent.<BR>\n" );
document.write( "since it is losing 11% of its value then it is growing by a rate of (1 minus .11) = .89 <BR>\n" );
document.write( "the formula becomes v(t) = 750 * .89 ^ t\r<BR>\n" );
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document.write( "you want to know how many years it will take for it to lose half its value.<BR>\n" );
document.write( "replace v(t) with 750/2 to get:<BR>\n" );
document.write( "750/2 = 750 * .89 ^ t<BR>\n" );
document.write( "if you divide both sides of that equation by 750, you will get:<BR>\n" );
document.write( "(750/2)/750 = .89 ^ t<BR>\n" );
document.write( "simplify that to get:<BR>\n" );
document.write( "375/750 = .89 ^ t<BR>\n" );
document.write( "simplify further to get:<BR>\n" );
document.write( ".5 = .89^t\r<BR>\n" );
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document.write( "to solve for t, take the log of both sides of the equation to get:<BR>\n" );
document.write( "log(.5) = log(.89^t)<BR>\n" );
document.write( "by one of the laws of logs, log(.89^t) = t * log(.89)<BR>\n" );
document.write( "equation becomes:<BR>\n" );
document.write( "log(.5) = t * log(.89<BR>\n" );
document.write( "solve for t to get:<BR>\n" );
document.write( "t = log(.5) / log(.89)<BR>\n" );
document.write( "use your calculator to get:<BR>\n" );
document.write( " = 5.948034681.\r<BR>\n" );
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document.write( "confirm by replacing t in our original equation with that value to get:<BR>\n" );
document.write( "375 = 750 * .89 ^ 5.948034681.<BR>\n" );
document.write( "evaluate to get:<BR>\n" );
document.write( "375 = 750 * .5 = 375.<BR>\n" );
document.write( "this confirms the solution is correct.\r<BR>\n" );
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document.write( "your solution is that t = 5.948034681\r<BR>\n" );
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document.write( "here's a reference on logarithm arithmetic.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"https://www.purplemath.com/modules/logrules.htm\" target = \"_blank\">https://www.purplemath.com/modules/logrules.htm</a>\r<BR>\n" );
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document.write( "there is a formula that i use that is more informative.<BR>\n" );
document.write( "the formula is f = p * (1 + r) ^ n<BR>\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the interest rate per time period<BR>\n" );
document.write( "(1 + r) is the growth rate per time period<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
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document.write( "your formula was v(t) = a * b ^ t<BR>\n" );
document.write( "your a corresponds to my p<BR>\n" );
document.write( "your b corresponds to my growth rate.\r<BR>\n" );
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document.write( "if you had used my formula, then:<BR>\n" );
document.write( "p = 750<BR>\n" );
document.write( "r = -.11<BR>\n" );
document.write( "1 + r = 1 - .11 = .89<BR>\n" );
document.write( "f = 375\r<BR>\n" );
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document.write( "the formula would have become 375 = 750 * .89 ^ n<BR>\n" );
document.write( "the growth rate is per year<BR>\n" );
document.write( "n is in years.<BR>\n" );
document.write( "your time periods are therefore in years.\r<BR>\n" );
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document.write( "you would have solve for n the same way.<BR>\n" );
document.write( "375 = 750 * .89 ^ n becomes:<BR>\n" );
document.write( "375/750 = .89 ^ n<BR>\n" );
document.write( "you would then take the log of both sides of the equation to get:<BR>\n" );
document.write( "log(375/750) = log(.89^n) which would become:<BR>\n" );
document.write( "log(375/750) = n * log(.89) which would become:<BR>\n" );
document.write( "n = log(375/750) / log(.89) which would get you the same answer.\r<BR>\n" );
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document.write( "hopefully this helps.<BR>\n" );
document.write( "let me know if you have any questions.\r<BR>\n" );
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document.write( "theo\r<BR>\n" );
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document.write( " </SPAN>\n" );
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