document.write( "Question 399711: 120. Global Warming. The increasing global temperature can be modeled by the function:
\n" ); document.write( " I = 0.1e^0.02t,
\n" ); document.write( "where I is the increase in global temperature in degrees Celsius since 1900, and t is the number of years since 1900 (NASA, www.science.nasa.gov).\r
\n" ); document.write( "\n" ); document.write( "A. How much warmer will it be in 2010 than it was in 1950?
\n" ); document.write( "B. In what year will the global temperature be at 4 degrees higher than the global temperature in 2000?
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #283162 by jsmallt9(3758)\"\" \"About 
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\"I+=+0.1e%5E0.02t\"
\n" ); document.write( "A. How much warmer will it be in 2010 than it was in 1950?
\n" ); document.write( "To find out how much warmer we will subtract:
\n" ); document.write( "the increase in global temperature between 2010 and 1900 minus the increase in global temperature between 1950 and 1900.
\n" ); document.write( "the increase in global temperature between 2010 and 1900: \"0.1e%5E%280.02%282010-1900%29%29+=+0.1e%5E%280.02%28110%29%29+=+0.1e%5E%282.2%29\"
\n" ); document.write( "the increase in global temperature between 1950 and 1900: \"0.1e%5E%280.02%281950-1900%29%29+=+0.1e%5E%280.02%2850%29%29+=+0.1e%5E1+=+0.1e\"
\n" ); document.write( "The difference will be
\n" ); document.write( "\"0.1e%5E%282.2%29+-+0.1e\"
\n" ); document.write( "Using 2.7182818284590451 (or some rounded off version of it) for e we get:
\n" ); document.write( "\"0.1%2A2.7182818284590451%5E%282.2%29+-+0.1%2A2.7182818284590451\"
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( "0.1*9.0250134994341199 - 0.1*2.7182818284590451
\n" ); document.write( "0.90250134994341199 - 0.27182818284590451
\n" ); document.write( "0.6306731670975075
\n" ); document.write( "So global temperature will be approximately 0.63 degrees Celsius warmer in 2010 than it was in 1950.

\n" ); document.write( "B. In what year will the global temperature be at 4 degrees higher than the global temperature in 2000?
\n" ); document.write( "Again we will use a difference. This time, however, we know what the difference is but we do not know what one of the years is:
\n" ); document.write( "\"4+=+0.1e%5E0.02t+-+0.1e%5E%280.02%282000-1900%29%29\"
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( "\"4+=+0.1e%5E0.02t+-+0.1e%5E%280.02%28100%29%29\"
\n" ); document.write( "\"4+=+0.1e%5E0.02t+-+0.1e%5E2\"
\n" ); document.write( "we will make things a little simpler if we multiply both sides by 10. (It will eliminate the 0.1's.)
\n" ); document.write( "\"40+=+e%5E0.02t+-+e%5E2\"
\n" ); document.write( "Now we can solve for t. First we add \"e%5E2\" to each side:
\n" ); document.write( "\"40+%2B+e%5E2+=+e%5E0.02t\"
\n" ); document.write( "Next, as is usually done when solving equations where the variable in in an exponent, we use logarithms. Since the base of the exponent is e, the natural choice (excuse the pun) is to use ln:
\n" ); document.write( "\"ln%2840+%2B+e%5E2%29+=+ln%28e%5E0.02t%29\"
\n" ); document.write( "Next we use a property of logarithms, \"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\", to move the exponent out in front of the logarithm. It is this very property that is the reason we use logarithms in the first place. Moving the exponent out in front puts the variable in a place where we can \"get at it\". Using this property on the right side we get:
\n" ); document.write( "\"ln%2840+%2B+e%5E2%29+=+%280.02t%29ln%28e%29\"
\n" ); document.write( "By definition ln(e) = 1 so this becomes:
\n" ); document.write( "\"ln%2840+%2B+e%5E2%29+=+0.02t\"
\n" ); document.write( "Dividing both sides by 0.02 we get:
\n" ); document.write( "\"ln%2840+%2B+e%5E2%29%2F0.02+=+t\"
\n" ); document.write( "This is an exact expression for the solution. For a decimal approximation we will use 2.7182818284590451 again for e:
\n" ); document.write( "\"ln%2840+%2B+2.7182818284590451%5E2%29%2F0.02+=+t\"
\n" ); document.write( "\"ln%2840+%2B+7.3890560989306494%29%2F0.02+=+t\"
\n" ); document.write( "\"ln%2847.3890560989306494%29%2F0.02+=+t\"
\n" ); document.write( "\"3.8583913180544145%2F0.02+=+t\"
\n" ); document.write( "192.9195659027207242 = t
\n" ); document.write( "So approximately 193 years after 1900, the year 2093, the temperature will be 4 degrees warmer than in 2000. \n" ); document.write( "
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