document.write( "Question 373981: The demand equation for a certain product is given by\r
\n" ); document.write( "\n" ); document.write( "p=5000(1-(4/(4+e^-0.002x)))\r
\n" ); document.write( "\n" ); document.write( "Find the demands x for prices of
\n" ); document.write( "a) p=$600
\n" ); document.write( "b) p=$400
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Algebra.Com's Answer #266278 by jsmallt9(3759)\"\" \"About 
You can put this solution on YOUR website!
\"p=5000%281-%284%2F%284%2Be%5E%28-0.002x%29%29%29%29\"
\n" ); document.write( "To find x for p = 600:
\n" ); document.write( "\"600+=+5000%281-%284%2F%284%2Be%5E%28-0.002x%29%29%29%29\"
\n" ); document.write( "We need \"peel away\" from the outside on the right side. First we get rid of the 5000 by dividing both sides be 5000:
\n" ); document.write( "\"600%2F5000+=+1-%284%2F%284%2Be%5E%28-0.002x%29%29%29\"
\n" ); document.write( "or
\n" ); document.write( "\"0.12+=+1-%284%2F%284%2Be%5E%28-0.002x%29%29%29\"
\n" ); document.write( "Next the 1 must go. Subtract 1 from each side:
\n" ); document.write( "\"-0.88+=+-%284%2F%284%2Be%5E%28-0.002x%29%29%29\"
\n" ); document.write( "Multiply (or divide) both sides by -1 to eliminate the minus sign:
\n" ); document.write( "\"0.88+=+4%2F%284%2Be%5E%28-0.002x%29%29\"
\n" ); document.write( "Next we'll eliminate the fraction by multiplying both sides by \"%284%2Be%5E%28-0.002x%29%29\":
\n" ); document.write( "\"0.88%29%2A%284%2Be%5E%28-0.002x%29%29+=+4\"
\n" ); document.write( "Using the Distributive Property on the left side:
\n" ); document.write( "\"0.88%29%2A%284%29%2B%280.88%29%2Ae%5E%28-0.002x%29+=+4\"
\n" ); document.write( "\"3.52+%2B+%280.88%29%2Ae%5E%28-0.002x%29+=+4\"
\n" ); document.write( "Subtracting 3.52 from each side:
\n" ); document.write( "\"%280.88%29%2Ae%5E%28-0.002x%29+=+0.48\"
\n" ); document.write( "Divide by 0.88:
\n" ); document.write( "\"e%5E%28-0.002x%29+=+%280.88%29%2A0.48\"
\n" ); document.write( "\"e%5E%28-0.002x%29+=+0.4224\"
\n" ); document.write( "Now that the exponential part of the equation is isolated, we use logarithms to proceed. Since the base of the exponent is e, it is best to use base e logarithms (aka lm):
\n" ); document.write( "\"ln%28e%5E%28-0.002x%29%29+=+ln%280.4224%29\"
\n" ); document.write( "On the left side we can 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 property which is the very reason we use logarithms. It gives us a way to get the variable out of the exponent.) Using the property on the left side we get:
\n" ); document.write( "\"%28-0.002x%29ln%28e%29+=+ln%280.4224%29\"
\n" ); document.write( "By definition, \"ln%28e%29+=+1\" so this becomes:
\n" ); document.write( "\"-0.002x+=+ln%280.4224%29\"
\n" ); document.write( "Last of all we divide by -0.002:
\n" ); document.write( "\"x+=+ln%280.4224%29%2F%28-0.002%29\"
\n" ); document.write( "This is an exact expression for your answer. Use your calculator if you want a decimal approximation:
\n" ); document.write( "\"x+=+%28-0.8618025465900853%29%2F%28-0.002%29\"
\n" ); document.write( "x = 430.9012732950426674
\n" ); document.write( "So for a price of $699 the demand will be approximately 431.

\n" ); document.write( "For a price of $400, replace the p with 400 and repeat these steps to find the demand at that price.
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