document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1209756>Question 1209756</A>: <I><SPAN STYLE=\"word-break: break-all;\"> . A construction company purchases a bulldozer for $160.000. Each year the value of the bulldozer depreciates by 20% of its value in the preceding year. Let V, be the value of the bulldozer in the nth year. (Let n = 1 be the year the bulldozer is purchased)\r<BR>\n" );
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document.write( "a) Find the formula for V.\r<BR>\n" );
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document.write( "12 Marks)\r<BR>\n" );
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document.write( "b) In what year will the value of the bulldozer be less than $100.000\r<BR>\n" );
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document.write( "(4 Marks) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850156 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><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=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850156\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this depreciation problem:\r<BR>\n" );
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document.write( "**a) Formula for V:**\r<BR>\n" );
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document.write( "* **Initial Value:** The bulldozer starts at $160,000.<BR>\n" );
document.write( "* **Depreciation Rate:** The bulldozer depreciates by 20% (0.20) each year.<BR>\n" );
document.write( "* **Value After 1 Year:** $160,000 * (1 - 0.20) = $160,000 * 0.80 = $128,000<BR>\n" );
document.write( "* **Value After 2 Years:** $128,000 * 0.80 = $160,000 * 0.80 * 0.80 = $160,000 * (0.80)²<BR>\n" );
document.write( "* **General Formula:** V(n) = $160,000 * (0.80)^(n-1)\r<BR>\n" );
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document.write( "**b) Year When Value is Less Than $100,000:**\r<BR>\n" );
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document.write( "We want to find the smallest integer *n* such that V(n) < $100,000.\r<BR>\n" );
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document.write( "1.  **Set up the inequality:**<BR>\n" );
document.write( "    $160,000 * (0.80)^(n-1) < $100,000\r<BR>\n" );
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document.write( "2.  **Divide by 160,000:**<BR>\n" );
document.write( "    (0.80)^(n-1) < 100,000 / 160,000<BR>\n" );
document.write( "    (0.80)^(n-1) < 0.625\r<BR>\n" );
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document.write( "3.  **Take the logarithm (base 10 or natural log) of both sides:**<BR>\n" );
document.write( "    log(0.80)^(n-1) < log(0.625)<BR>\n" );
document.write( "    (n-1) * log(0.80) < log(0.625)\r<BR>\n" );
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document.write( "4.  **Divide by log(0.80). Note that log(0.80) is negative, so reverse the inequality sign:**<BR>\n" );
document.write( "    n - 1 > log(0.625) / log(0.80)<BR>\n" );
document.write( "    n - 1 > -0.2076 / -0.0969<BR>\n" );
document.write( "    n - 1 > 2.1424\r<BR>\n" );
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document.write( "5.  **Add 1 to both sides:**<BR>\n" );
document.write( "    n > 3.1424\r<BR>\n" );
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document.write( "6.  **Find the smallest integer:**<BR>\n" );
document.write( "    Since *n* must be an integer, the smallest value of *n* that satisfies this inequality is *n* = 4.\r<BR>\n" );
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document.write( "**Answer:**\r<BR>\n" );
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document.write( "a) The formula for V is V(n) = $160,000 * (0.80)^(n-1)\r<BR>\n" );
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document.write( "b) The value of the bulldozer will be less than $100,000 in the 4th year.<BR>\n" );
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