document.write( "Question 1153619: The population P of a fish farm in t years is modeled by the equation
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\n" ); document.write( "To the nearest tenth, how long will it take for the population to reach 900?
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Algebra.Com's Answer #775874 by MathLover1(20850)\"\" \"About 
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The population P of a fish farm in t years is modeled by the equation:\r
\n" ); document.write( "\n" ); document.write( "\"P%28t%29+=2000%2F%281+%2B+9e%5E%28-0.6t%29%29\"\r
\n" ); document.write( "\n" ); document.write( "To the nearest tenth, how long will it take for the population to reach \"900\"?\r
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\n" ); document.write( "\n" ); document.write( "find \"P%28t%29+=900\"\r
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\n" ); document.write( "\n" ); document.write( "\"900+=2000%2F%281+%2B+9e%5E%28-0.6t%29%29\"........multiply both sides by \"%281+%2B+9e%5E%28-0.6t%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"900%2A%281%2B9e%5E%28-0.6t%29%29+=2000\"......divide both sides by \"900\"\r
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\n" ); document.write( "\n" ); document.write( "\"1%2B9e%5E%28-0.6t%29=20%2F9\"...move 1 to the right side\r
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\n" ); document.write( "\n" ); document.write( "\"9e%5E%28-0.6t%29=20%2F9-1\"\r
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\n" ); document.write( "\n" ); document.write( "\"9e%5E%28-0.6t%29=20%2F9-9%2F9\"\r
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\n" ); document.write( "\n" ); document.write( "\"9e%5E%28-0.6t%29=11%2F9\"\r
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\n" ); document.write( "\n" ); document.write( "\"e%5E%28-0.6t%29=%2811%2F9%29%2F9\"\r
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\n" ); document.write( "\n" ); document.write( "\"e%5E%28-0.6t%29=11%2F81\"....take natural logarithm of both sides\r
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\n" ); document.write( "\n" ); document.write( "\"ln%28e%5E%28-0.6t%29%29=ln%2811%2F81%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28-0.6t%29+ln%28e%29=ln%2811%29-ln%2881%29\".....\"ln%28e%29=1\"\r
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\n" ); document.write( "\n" ); document.write( "\"-0.6t=ln%2811%29-ln%2881%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"t=%28ln%2811%29-ln%2881%29%29%2F-0.6\"\r
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\n" ); document.write( "\n" ); document.write( "\"t\"\"3.32759+\"\r
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\n" ); document.write( "\n" ); document.write( "To the nearest tenth, it will take \"3.3\" for the population to reach \"900\"\r
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