document.write( "Question 420210: The value of a country's exports for some product(in billions of dollars) is approximated by the function f(x)=1.3e to the power of .2385x, where x=0 corresponds to the year 1995. What was the value of this country's exports in the year 2000?
\n" ); document.write( "Using the information from that problem, in what year did the level of exports reach 15 billions? \n" ); document.write( "

Algebra.Com's Answer #293711 by jsmallt9(3758) $\"\"$ $\"About$

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$\"f%28x%29+=+1.3%2Ae%5E%280.2385x%29\"$
\n" ); document.write( "For the first part, if x = 0 in 1995 then x = 5 in 2000. To find the exports in 2000, we find f(5):
\n" ); document.write( " $\"f%285%29+=+1.3%2Ae%5E%280.2385%285%29%29\"$
\n" ); document.write( "Just enter this into your calculator to find the answer. (If your calculator does not have a button for \"e\", then use 2.7182818284590451 (or some rounded of version of it).)

\n" ); document.write( "For the second part we are interested in when the exports reach 15 billion. In mathematical terms we are interested the x that makes f(x) a 15:
\n" ); document.write( " $\"15+=+1.3%2Ae%5E%280.2385x%29\"$
\n" ); document.write( "To solve this for x we start by isolating the base and its exponent. Dividing both sides by 1.3 we get:
\n" ); document.write( " $\"11.5384615384615385+=+e%5E%280.2385x%29\"$
\n" ); document.write( "Next we use logarithms. Any base of logarithm can be used. But if we match the base of the logarithm with the base of the exponent we will end up with a simpler expression. So we will use base e logarithms (aka ln):
\n" ); document.write( " $\"ln%2811.5384615384615385%29+=+ln%28e%5E%280.2385x%29%29\"$
\n" ); document.write( "Next we use a property of logarithms, $\"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\"$ , which allows us to move the exponent of an argument out in front of the logarithm. (It is this very property that is the reason we use logarithms on equations like this. The property allows us to move the exponent, where the variable is, to a location where we can then solve for the variable. And we can use any base of logarithm since this property works with all logarithms.) Using this property on our equation we get:
\n" ); document.write( " $\"ln%2811.5384615384615385%29+=+%280.2385x%29%2Aln%28e%29\"$
\n" ); document.write( "Since ln(e) = 1 (which is why matching the bases gives us a simpler expression) this becomes:
\n" ); document.write( " $\"ln%2811.5384615384615385%29+=+0.2385x\"$
\n" ); document.write( "Now that x is out of the exponent we can solve for it. Dividing both sides by 0.2385:
\n" ); document.write( " $\"ln%2811.5384615384615385%29%2F0.2385+=+x\"$
\n" ); document.write( "This is an exact expression for the solution. But it doesn't help us say what year the exports reach 15 billion. So we need a decimal approximation. Using our calculators we get:
\n" ); document.write( " $\"2.4456859366347190%2F0.2385+=+x\"$
\n" ); document.write( "10.2544483716340420 = x
\n" ); document.write( "Since x = 0 is 1995, this answer tells us that approximately 10 years later, 2005, the exports will reach 15 billion. \n" ); document.write( "

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