document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1159949>Question 1159949</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello<BR>\n" );
document.write( "The answer to the following question is 26 but I'm trying to figure out how to set up the equation and enter it into the calculator with x being the exponent.\r<BR>\n" );
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document.write( "Appreciate your help on this one. Thanks\r<BR>\n" );
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document.write( "Picture a cat stalking a mouse. They’re about 100 inches apart. Every time the mouse starts nibbling at the hunk of cheese, the cat takes advantage of the mouse’s distraction and creeps closer by one-tenth the distance between them. The cat want's to get about 6 inches away. How long will it take before the cat can pounce on the mouse. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #783078 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=783078\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "x = number of nibbles the mouse makes<BR>\n" );
document.write( "y = distance between the cat and mouse<BR>\n" );
document.write( "x is a positive whole number, while y is a positive real number<BR>\n" );
document.write( "x cannot be a fraction or decimal, but y can\r<BR>\n" );
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document.write( "When x = 0, the value of y is y = 100. This is the starting distance.\r<BR>\n" );
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document.write( "When x changes to x = 1, the value of y becomes y = 90. We take 1/10 of 100 to get 10, so the cat has moved 10 inches closer leaving 100-10 = 90 inches left til the cat reaches the mouse.\r<BR>\n" );
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document.write( "When we get to x = 2, the cat moves (1/10)*90 = 90/10 = 9 inches. So there's 90-9 = 81 inches left.\r<BR>\n" );
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document.write( "You can keep this going until you have y be 6 inches. If you can't land exactly on y = 6, then try to get as close as possible. Make sure to have y be smaller than 6, rather than larger, so that we meet our goal.\r<BR>\n" );
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document.write( "As you can probably see, it will take a while to get y to 6 or smaller following the method outlined above. We can use algebra to help speed things up.\r<BR>\n" );
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document.write( "Notice that each time the cat moves 1/10 of the distance closer, there's 9/10 of the distance left to go. What we can do is multiply 9/10 by the old distance remaining to get the new distance remaining\r<BR>\n" );
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document.write( "new distance remaining = (9/10)*(old distance remaining)\r<BR>\n" );
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document.write( "So for instance, we start off with 100 and go to 90<BR>\n" );
document.write( "new distance remaining = (9/10)*(old distance remaining)<BR>\n" );
document.write( "new distance remaining = (9/10)*(100)<BR>\n" );
document.write( "new distance remaining = 90\r<BR>\n" );
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document.write( "and this repeats for the next iteration as well<BR>\n" );
document.write( "new distance remaining = (9/10)*(old distance remaining)<BR>\n" );
document.write( "new distance remaining = (9/10)*(90)<BR>\n" );
document.write( "new distance remaining = 81\r<BR>\n" );
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document.write( "and so on. When we have repeated multiplication, we turn to exponents. You probably already know this considering you mentioned exponents in your post.\r<BR>\n" );
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document.write( "The equation would be<BR>\n" );
document.write( "y = 100(9/10)^x<BR>\n" );
document.write( "which in decimal form is<BR>\n" );
document.write( "y = 100(0.9)^x\r<BR>\n" );
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document.write( "Let's try x = 0 and see what happens:<BR>\n" );
document.write( "y = 100(0.9)^x<BR>\n" );
document.write( "y = 100(0.9)^0<BR>\n" );
document.write( "y = 100(1)<BR>\n" );
document.write( "y = 100<BR>\n" );
document.write( "so we have x = 0 and y = 100 pair up as expected\r<BR>\n" );
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document.write( "Now try x = 1<BR>\n" );
document.write( "y = 100(0.9)^x<BR>\n" );
document.write( "y = 100(0.9)^1<BR>\n" );
document.write( "y = 100(0.9)<BR>\n" );
document.write( "y = 90<BR>\n" );
document.write( "That works as well. The cat is y = 90 inches away after the mouse does x = 1 nibble.\r<BR>\n" );
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document.write( "Lastly lets try out x = 2<BR>\n" );
document.write( "y = 100(0.9)^x<BR>\n" );
document.write( "y = 100(0.9)^2<BR>\n" );
document.write( "y = 100(0.81)<BR>\n" );
document.write( "y = 81<BR>\n" );
document.write( "This works too.\r<BR>\n" );
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document.write( "We want to solve the exponential equation for x when y = 6<BR>\n" );
document.write( "We will use logarithms to help isolate the exponent. One way you can remember is thinking \"the exponent is in the trees, so we have to log it down\"\r<BR>\n" );
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document.write( "Let's do so<BR>\n" );
document.write( "y = 100(0.9)^x<BR>\n" );
document.write( "6 = 100(0.9)^x<BR>\n" );
document.write( "100(0.9)^x = 6<BR>\n" );
document.write( "(0.9)^x = 6/100<BR>\n" );
document.write( "(0.9)^x = 0.06<BR>\n" );
document.write( "Log[ (0.9)^x ] = Log[0.06] .... apply logs to both sides<BR>\n" );
document.write( "x*Log[0.9] = Log[0.06] .... see note below<BR>\n" );
document.write( "x = Log[0.06]/Log[0.9]<BR>\n" );
document.write( "x = 26.7027045112137\r<BR>\n" );
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document.write( "note: I'm using the log rule log(x^y) = y*log(x), which is the sole reason why we use logarithms. This rule allows us to pull the exponent down.\r<BR>\n" );
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document.write( "If x = 26, then,<BR>\n" );
document.write( "y = 100(0.9)^x<BR>\n" );
document.write( "y = 100(0.9)^26<BR>\n" );
document.write( "y = 100(0.06461081889227)<BR>\n" );
document.write( "y = 6.46108188922668<BR>\n" );
document.write( "which rounds to y = 6 when we round to the nearest whole number. \r<BR>\n" );
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document.write( "I would argue that x = 26 nibbles is 1 too few because y = 6.46 is over 6 inches. The cat would probably want to get under y = 6 (if the cat can't exactly land on y = 6 itself) so I would argue x = 27 is a better fit.\r<BR>\n" );
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document.write( "y = 100(0.9)^x<BR>\n" );
document.write( "y = 100(0.9)^27 ... try out x = 27<BR>\n" );
document.write( "y = 100(0.05814973700304)<BR>\n" );
document.write( "y = 5.814973700304<BR>\n" );
document.write( "which is now under 6\r<BR>\n" );
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document.write( "However, I think I see why your teacher opted to go with x = 26 instead of x = 27.<BR>\n" );
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