document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1034586>Question 1034586</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A starting salary for a teacher is $25,000 and there is an annual increase of 3%.\r<BR>\n" );
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document.write( "(a) how much will the teacher earn in their 10th year?\r<BR>\n" );
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document.write( "(b)  how much will the teacher earn in total during a 35 year teaching career? \r<BR>\n" );
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document.write( "(c) find the first year in which the teacher earns more than $35,000?\r<BR>\n" );
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document.write( "(d)  how many years would the teacher have to work in order to earn a total of $1 million?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #649278 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=649278\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> this looks like a geometric progression where the common ratio is 1.03.\r<BR>\n" );
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document.write( "the nth term of a geometric progression is given by the formula:\r<BR>\n" );
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document.write( "An = A1 * r^(n-1).\r<BR>\n" );
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document.write( "when A1 = 25,000, and r = 1.03, the formula becomes:\r<BR>\n" );
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document.write( "An = 25,000 * 1.03^(n-1).\r<BR>\n" );
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document.write( "when n = 1, you get A1 = 25,000 * 1.03^(0) = 25,000.<BR>\n" );
document.write( "when n = 2, you get A2 = 25,000 * 1.03^1 = 25,750.\r<BR>\n" );
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document.write( "the sum of a geometric progression is given by the formula:\r<BR>\n" );
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document.write( "Sn = A1 * (1 - r^n) / (1-r)\r<BR>\n" );
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document.write( "when n = 1, you get S1 = 25,000 * (1 - 1.03^1) / (1-1.03) = 25,000<BR>\n" );
document.write( "when n = 2, you get S2 = 25,000 * (1 - 1.03^2) / (1 - 1.03) = 50,750.\r<BR>\n" );
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document.write( "since we know what A1 and A2 are from above, we can just add them together to get 25,000 + 25,750 = 50,750.\r<BR>\n" );
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document.write( "since this is the same as the Sn formula when n = 2, we have just confirmed that the formulas are good.\r<BR>\n" );
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document.write( "now we can answer the questions you have.\r<BR>\n" );
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document.write( "the questions with their answers are shown below:\r<BR>\n" );
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document.write( "(a) how much will the teacher earn in their 10th year?\r<BR>\n" );
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document.write( "A10 = A1 * r^(n-1) becomes:<BR>\n" );
document.write( "A10 = 25,000 * 1.03^9 which becomes:<BR>\n" );
document.write( "A10 = 32,619.3296<BR>\n" );
document.write( "the teacher will have earned 32,619.3296 by the end of their 10th year.\r<BR>\n" );
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document.write( "(b) how much will the teacher earn in total during a 35 year teaching career?\r<BR>\n" );
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document.write( "S35 = A1 * (1 - r^35) / (1-r) becomes:<BR>\n" );
document.write( "S35 = 25,000 * (1 - 1.03^35) / (1-1.03) which becomes:<BR>\n" );
document.write( "S35 = 1,511,552.045<BR>\n" );
document.write( "the teacher will earn a total of 1,511,552.045 by the end of their 35th year.\r<BR>\n" );
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document.write( "(c) find the first year in which the teacher earns more than $35,000?\r<BR>\n" );
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document.write( "An = A1 * r^(n-1) becomes:<BR>\n" );
document.write( "35,000 = 25,000 * 1.03^(n-1).<BR>\n" );
document.write( "divide both sides of this equation by 25,000 to get:<BR>\n" );
document.write( "35,000 / 25,000 = 1.03^(n-1)<BR>\n" );
document.write( "take the log of both sides of this equation to get:<BR>\n" );
document.write( "log(35,000/25,000) = log(1.03^(n-1)<BR>\n" );
document.write( "since log(a^b) = b*log(a), this equation becomes:<BR>\n" );
document.write( "log(35,000/25,000 + (n-1) * log(1.03)<BR>\n" );
document.write( "divide both sides of this equation by log(1.03) to get:<BR>\n" );
document.write( "log(35,000/25,000) / log(1.03) = n-1.<BR>\n" );
document.write( "add 1 to both sides of this equation to get:<BR>\n" );
document.write( "log(35,000/25,000) / log(1.03) + 1 = n<BR>\n" );
document.write( "solve for n to get n = log(35,000/25,000) / log(1.03) + 1 = 12.38314854<BR>\n" );
document.write( "confirm by replacing n in the equation of 35,000 = 25,000 * 1.03^(n-1) to get:<BR>\n" );
document.write( "35,000 = 25,000 * 1.03^(12.38314854-1).<BR>\n" );
document.write( "you will get 35,000 = 35,000.<BR>\n" );
document.write( "when n = 12, the teacher earns 25,000 * 1.03^11 = 34,605.85<BR>\n" );
document.write( "this means the teacher earns 34,605.85 by the end of the 12th year.<BR>\n" );
document.write( "when n = 13, the teacher earns 25,000 * 1.03^(12 = 35,644.02.<BR>\n" );
document.write( "this means the teacher earns 35,644.01 by the end of the 13th year.<BR>\n" );
document.write( "this means that the first year that the teacher earns 35,000 is sometime between the end of the 12th year and the end of the 13th year.<BR>\n" );
document.write( "this puts the first year that the teacher first earns 35,000 in the 13th year.\r<BR>\n" );
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document.write( "(d) how many years would the teacher have to work in order to earn a total of $1 million? \r<BR>\n" );
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document.write( "Sn = A1 * (1-r^n) / (1-r) becomes:<BR>\n" );
document.write( "1,000,000 = 25,000 * (1-1.03^n) / (1-1.03)<BR>\n" );
document.write( "divide both sides of this equation by 25,000 to get:<BR>\n" );
document.write( "1,000,000/25,000 = (1-1.03^n) / (1-1.03)<BR>\n" );
document.write( "multiply both sides of this equation by (1-1.03) to get:<BR>\n" );
document.write( "(1,000,000/25,000) * (1-1.03) = 1 - 1.03^n<BR>\n" );
document.write( "simplify to get:<BR>\n" );
document.write( "-1.2 = 1 - 1.03^n.<BR>\n" );
document.write( "subtract 1 from both sides of this equation to get:<BR>\n" );
document.write( "-2.2 = -1.03^n<BR>\n" );
document.write( "multiply both sides of this equation by -1 to get:<BR>\n" );
document.write( "2.2 = 1.03^n<BR>\n" );
document.write( "take the log of both sides of this equation to get:<BR>\n" );
document.write( "log(2.2) = log(1.03^n)<BR>\n" );
document.write( "since log(a^b) = b*log(a), this equation becomes:<BR>\n" );
document.write( "log(2.2) = n * log(1.03)<BR>\n" );
document.write( "divide both sides of this equaiton by log(1.03) to get:<BR>\n" );
document.write( "log(2.2) / log(1.03) = n<BR>\n" );
document.write( "solve for n to get n = log(2.2) / log(1.03) = 26.67419857.<BR>\n" );
document.write( "replace n in the original equation of 1,000,000 = 25,000 * (1-1.03^n) / (1-1.03) to get:<BR>\n" );
document.write( "1,000,000 = 25,000 * (1-1.03^26.67419857) / (1-1.03) = 1,000,000.<BR>\n" );
document.write( "this confirms the solution is correct.<BR>\n" );
document.write( "when n = 26, the teacher earns a total of S26 = 25,000 * (1 - 1.03^26)/(1-1.03) = 96,3826.06.<BR>\n" );
document.write( "this means the teacher earns a total of 96,3826.06 by the end of the 26th year.<BR>\n" );
document.write( "when n = 27, the teacher earns a total of S27 = 25,000 * (1-1.03^27)/(1-1.03) = 1,017,740.84.<BR>\n" );
document.write( "this means the teacher earns a total of 1,017,740.84 by the end of the 27th year.<BR>\n" );
document.write( "this means that the first year that the teacher earns 1,000,000 is sometime between the end of the 26th year and the end of the 27th year.<BR>\n" );
document.write( "this puts the first year that the teacher first earns 1,000,000 in the 27th year.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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