SOLUTION: The demand function for a certain commodity is given by p=80e exponent fraction -q/2 Write q as a function of p

Question 70663: The demand function for a certain commodity is given by p=80e exponent fraction -q/2 Write q as a function of p
Found 2 solutions by funmath, bucky:
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
The demand function for a certain commodity is given by p=80e exponent fraction -q/2 Write q as a function of p

ln(e)=1, so

Happy Calculating!!!!
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Solve for q.
.
This is what I understand your problem to be. Assuming this is correct, you can take the ln
(natural logarithm which has the base e) of both sides and the problem becomes:
.

.
But the logarithm of a product is equals the sum of the logarithms of the two terms being
multiplied. Therefore, we can split the right side into the sum of two logarithms as follows:
.

.
Subtract ln(80) from both sides to get:
.

.
By the rules of logarithms, the difference of the logarithms of two quantities can be
re-written as the logarithm of the quotients of the quantities. This translates to:
.

.
Substituting this as a replacement for the left side results in:
.

.
Then by a rule of exponents in logarithms, the exponent of a term in a logarithm becomes the
multiplier of the logarithm of the term on the right side. In this case
becomes the multiplier of ln(e) and the right side of the equation is changed as shown below:
.

.
But ln(e) = 1, and when this substitution is made the equation becomes:
.

.
Multiply both sides of the equation by -2 and the equation becomes:
.

.
This is the answer, but there is one additional constraint. The value of p must be greater
than zero or else you would be taking the ln of a negative number or zero and those are
outside of the allowed values of numbers that the ln function can operate on.
.
Hopes this gives you some additional insight about the subject of logarithms and natural
logarithms in particular.
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