document.write( "Question 240616: P=13500+22ln(t+1), where t is th etime in years from the present. In how many years will there be 18,000 omsects?(round to the nearest tenth of a year).\r
\n" ); document.write( "\n" ); document.write( "18,000=13,500+22ln(t+1)
\n" ); document.write( "4,500=22ln(t+1)
\n" ); document.write( "204.5=ln(t+1)
\n" ); document.write( "

Algebra.Com's Answer #176281 by rapaljer(4671)\"\" \"About 
You can put this solution on YOUR website!
You are correct so far. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "18,000=13,500+22ln(t+1)
\n" ); document.write( "4,500=22ln(t+1)
\n" ); document.write( "204.5=ln(t+1)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The next step is to \"undo\" the ln function. You can do this by raising both sides as a power of \"e\".\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"e%5E204.5=+e%5E%28ln%28t%2B1%29%29\"
\n" ); document.write( "\"e%5E204.5=+t%2B1\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Now, subtract 1 from each side:
\n" ); document.write( "\"t=e%5E204.5-1\", which is approximately 6.5046*10^88 years, which is one HECK of a long time!! Are you sure you copied this problem correctly?? Are all the units given in years???\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Dr. Robert J. Rapalje, Retired
\n" ); document.write( "Seminole State College of Florida
\n" ); document.write( "Altamonte Springs Campus
\n" ); document.write( "
\n" );