document.write( "Question 377930: Solve for t. 6200e^-0.08t = 1400 \n" ); document.write( "

Algebra.Com's Answer #268639 by jsmallt9(3758) $\"\"$ $\"About$

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$\"6200e%5E%28-0.08t%29+=+1400\"$
\n" ); document.write( "Although not required, it is often easier to eliminate the coefficient in front of the base and its exponent. So we will eliminate the 6200 by dividing both sides by 6200:
\n" ); document.write( " $\"e%5E%28-0.08t%29+=+1400%2F6200\"$
\n" ); document.write( "The fraction will reduce:
\n" ); document.write( " $\"e%5E%28-0.08t%29+=+14%2F62\"$
\n" ); document.write( " $\"e%5E%28-0.08t%29+=+7%2F31\"$

\n" ); document.write( "Now, as is often done with equations where the variable is in an exponent, we will use logarithms. (We could have started with this but, as mentioned above, eliminating the 6200 in front makes the rest of the problem a little easier. Later I will show you how it looks when you don't elimimate the coefficient.) The base of the logarithm we use is not especially important. But if we use a logarithm with the same base as the base with the exponent, in this case it is e, then our answer will be a simpler expression. So we will use base e (or ln) logarithms. (I will show use of logarithms of a different base later.)
\n" ); document.write( " $\"ln%28e%5E%28-0.08t%29%29+=+ln%287%2F31%29\"$
\n" ); document.write( "On the left side we use a property of logarithms, $\"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\"$ , to \"move\" the exponent of the argument out in front. (It is this very property that is the reason we use logarithms on these equations. It allows us to move the exponent, where the variable is, to a place we we \"can get at it\" with Algebra.) Using this property on the left side we get:
\n" ); document.write( " $\"%28-0.08t%29%2Aln%28e%29+=+ln%287%2F31%29\"$
\n" ); document.write( "By definition, ln(e) = 1. (This is why choosing the base of the logarithm to match the base with the exponent makes the answer simpler.) So this becomes:
\n" ); document.write( " $\"-0.08t+=+ln%287%2F31%29\"$
\n" ); document.write( "Now we just divide both sides by -0.08:
\n" ); document.write( " $\"t+=+ln%287%2F31%29%2F-0.08\"$
\n" ); document.write( "This is an exact expression for the answer. If you need a decimal approximation, then get out your calculator, find the ln(7/31) and then divide it by -0.08.

\n" ); document.write( "If you don't eliminate the 6200 first...
\n" ); document.write( " $\"ln%286200e%5E%28-0.08t%29%29+=+ln%281400%29\"$
\n" ); document.write( "This time, because the 6200 is still there and because the exponent on e does not apply to the 6200, we must separate the 6200 before we use the property to move the exponent out in front. Fortunately there is another property of logarithms, $\"log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28q%29%29\"$ , which allows us to separate the logarithm of a product into the sum of the logarithms fo the factors. Using this on the left side we get:
\n" ); document.write( " $\"ln%286200%29+%2B+ln%28e%5E%28-0.08t%29%29+=+ln%281400%29\"$
\n" ); document.write( "Now we can use the property to move the exponent on the second logarithm on the left side:
\n" ); document.write( " $\"ln%286200%29+%2B+%28-0.08t%29%2Aln%28e%29+=+ln%281400%29\"$
\n" ); document.write( "And again ln(e) = 1 so this becomes:
\n" ); document.write( " $\"ln%286200%29+%2B+%28-0.08t%29+=+ln%281400%29\"$
\n" ); document.write( "Now we solve for t. First we'll subtract ln(6200) from each side:
\n" ); document.write( " $\"-0.08t+=+ln%281400%29+-+ln%286200%29\"$
\n" ); document.write( "And finally we divide by -0.08:
\n" ); document.write( " $\"t+=+%28ln%281400%29+-+ln%286200%29%29%2F-0.08\"$
\n" ); document.write( "This is another exact expression for the solution. If we use calculators on it to find a decimal approximation it should work out the same as (or extremely close to) the decimal approximation of our earlier solution, $\"t+=+ln%287%2F31%29%2F-0.08\"$ .

\n" ); document.write( "Using logarithms of a different base...
\n" ); document.write( "Let's say we decide to use base 10 logarithms instead. Then we get:
\n" ); document.write( " $\"log%28%28e%5E%28-0.08t%29%29%29+=+log%28%287%2F31%29%29\"$
\n" ); document.write( "Using the property again to move the exponent we get:
\n" ); document.write( " $\"%28-0.08t%29%2Alog%28%28e%29%29+=+log%28%287%2F31%29%29\"$
\n" ); document.write( "Factoring out t on the left side we get:
\n" ); document.write( " $\"t%28%28-0.08%29%2Alog%28%28e%29%29%29+=+log%28%287%2F31%29%29\"$
\n" ); document.write( "And then divide both sides by ((-0.08)*log((e))):
\n" ); document.write( " $\"t+=+log%28%287%2F31%29%29%2F%28%28-0.08%29%2Alog%28%28e%29%29%29\"$
\n" ); document.write( "This is yet another exact expression for the solution to this equation. It is a little more complex than $\"t+=+ln%287%2F31%29%2F-0.08\"$ but this is because we were not able to simplify log(e) like we were able to do with ln(e). For a decimal approximation we need our calculators. The decimal approximation here should be the same as (or extremely close to) the ones we would get from either of the other solutions we have found, $\"t+=+ln%287%2F31%29%2F-0.08\"$ or $\"t+=+%28ln%281400%29+-+ln%286200%29%29%2F-0.08\"$ . \n" ); document.write( "

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