document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1169051>Question 1169051</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The sample of work produced by lathe is 10000.<BR>\n" );
document.write( "a. If the machine doubled this production every 20 minutes, how much will it produce<BR>\n" );
document.write( "in one hour?<BR>\n" );
document.write( "b. How long will it take the machine to produce 8 million? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851455 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=851455\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the initial production of the lathe be $P_0 = 10000$.\r<BR>\n" );
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document.write( "**a. Production in one hour:**\r<BR>\n" );
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document.write( "One hour has 60 minutes. The machine doubles its production every 20 minutes. So, in one hour, the production will double $60 / 20 = 3$ times.\r<BR>\n" );
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document.write( "* After the first 20 minutes: $P_1 = P_0 \times 2 = 10000 \times 2 = 20000$<BR>\n" );
document.write( "* After the next 20 minutes (total 40 minutes): $P_2 = P_1 \times 2 = 20000 \times 2 = 40000$<BR>\n" );
document.write( "* After the final 20 minutes (total 60 minutes or one hour): $P_3 = P_2 \times 2 = 40000 \times 2 = 80000$\r<BR>\n" );
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document.write( "Alternatively, we can use the formula:<BR>\n" );
document.write( "$P(t) = P_0 \times 2^{t/d}$<BR>\n" );
document.write( "where:<BR>\n" );
document.write( "* $P(t)$ is the production after time $t$<BR>\n" );
document.write( "* $P_0$ is the initial production<BR>\n" );
document.write( "* $t$ is the total time<BR>\n" );
document.write( "* $d$ is the doubling time\r<BR>\n" );
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document.write( "For one hour (60 minutes):<BR>\n" );
document.write( "$P(60) = 10000 \times 2^{60/20}$<BR>\n" );
document.write( "$P(60) = 10000 \times 2^3$<BR>\n" );
document.write( "$P(60) = 10000 \times 8$<BR>\n" );
document.write( "$P(60) = 80000$\r<BR>\n" );
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document.write( "So, the machine will produce **80,000** units in one hour.\r<BR>\n" );
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document.write( "**b. Time to produce 8 million:**\r<BR>\n" );
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document.write( "We want to find the time $t$ when the production $P(t)$ reaches 8,000,000.<BR>\n" );
document.write( "$P(t) = 10000 \times 2^{t/20} = 8000000$\r<BR>\n" );
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document.write( "Divide both sides by 10000:<BR>\n" );
document.write( "$2^{t/20} = \frac{8000000}{10000}$<BR>\n" );
document.write( "$2^{t/20} = 800$\r<BR>\n" );
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document.write( "To solve for $t$, we can take the logarithm of both sides (using base 2 or natural logarithm):\r<BR>\n" );
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document.write( "Using base 2 logarithm:<BR>\n" );
document.write( "$\log_2(2^{t/20}) = \log_2(800)$<BR>\n" );
document.write( "$\frac{t}{20} = \log_2(800)$\r<BR>\n" );
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document.write( "We know that $2^9 = 512$ and $2^{10} = 1024$. So, $\log_2(800)$ is between 9 and 10.<BR>\n" );
document.write( "$\log_2(800) = \log_2(8 \times 100) = \log_2(2^3 \times 100) = 3 + \log_2(100)$<BR>\n" );
document.write( "Since $2^6 = 64$ and $2^7 = 128$, $\log_2(100)$ is between 6 and 7 (approximately 6.64).<BR>\n" );
document.write( "$\log_2(800) \approx 3 + 6.64 = 9.64$\r<BR>\n" );
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document.write( "Now, solve for $t$:<BR>\n" );
document.write( "$t = 20 \times \log_2(800)$<BR>\n" );
document.write( "$t \approx 20 \times 9.64$<BR>\n" );
document.write( "$t \approx 192.8$ minutes\r<BR>\n" );
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document.write( "Alternatively, using natural logarithm:<BR>\n" );
document.write( "$\ln(2^{t/20}) = \ln(800)$<BR>\n" );
document.write( "$\frac{t}{20} \ln(2) = \ln(800)$<BR>\n" );
document.write( "$t = 20 \times \frac{\ln(800)}{\ln(2)}$<BR>\n" );
document.write( "$t \approx 20 \times \frac{6.6846}{0.6931}$<BR>\n" );
document.write( "$t \approx 20 \times 9.644$<BR>\n" );
document.write( "$t \approx 192.88$ minutes\r<BR>\n" );
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document.write( "So, it will take approximately **192.88 minutes** for the machine to produce 8 million units.\r<BR>\n" );
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document.write( "To express this in hours and minutes:<BR>\n" );
document.write( "$192.88 \text{ minutes} = 3 \text{ hours and } 12.88 \text{ minutes}$<BR>\n" );
document.write( "$0.88 \text{ minutes} \times 60 \text{ seconds/minute} \approx 53 \text{ seconds}$\r<BR>\n" );
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document.write( "So, it will take approximately 3 hours, 12 minutes, and 53 seconds.\r<BR>\n" );
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document.write( "Final Answers:<BR>\n" );
document.write( "a. The machine will produce **80,000** units in one hour.<BR>\n" );
document.write( "b. It will take approximately **192.88 minutes** (or about 3 hours, 12 minutes, and 53 seconds) for the machine to produce 8 million units. </SPAN>\n" );
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