MAT01 Mathematics for Management and Economics

Let monthly energy spending A (in CHF per month)  of a household as a function of its total income Y (in CHF per month) follow the assignment

\(A(Y) = 80 \cdot \ln (Y + 100) - 500 \quad \quad Y \ge 1000\)

a) Determine the assignment of the inverse function Y(A).

b) What is the total income corresponding to a monthly energy spending of 250 CHF?

c) For which total income is energy spending 3% of that income? (Use the calculator)?

Preliminary remarks

• The theoretical background of inverse functions is in block I, learning sequence 3 of the script.
• If necessary, have a look at the sections on percentages of the preparatory course.

a) Inverse function

To determine the inverse Y(A) of the function A(Y), solve the equation of A(Y) for Y, i.e. isolate this variable, so that the equation shows how different values of A determine the values of Y.

Thus, our starting point is the equation

\(A(Y) = 80 \cdot \ln (Y + 100) - 500\)

Hint: in cases like this, it makes sense to replace A(Y) by A. This simplifies terms and facilitates rearranging them.

\(A = 80 \cdot \ln (Y + 100) - 500\)

One of the terms in the right side part of the equation is ln, the natural logarithm. We know that the e-function, that is the exponential function with Eulers number e as a basis, is its inverse function. After some rearrangement, we put this function to work.

First, we add 500 to both sides

\(A+500 = 80 \cdot \ln (Y + 100) \)

Then, we divide both sides by 80

\(\frac{{A + 500}}{{80}} = \ln (Y + 100)\)

The next step consists of applying the e-function to both sides of the equation. The effect on the right side term is obvious:

\(e^{\ln(Y+100)}= Y + 100\), because the inverse function of the ln-function neutralizes the ln-function.

Thus we obtain

\({e^{\frac{{A + 500}}{{80}}}} = Y + 100\),

Subtract 100 from both sides to see that

\(Y = {e^{\frac{{A + 500}}{{80}}}} - 100 \)

Now that the work has been done, we may add the independent variable A to the left side of the equation. Replacing Y by Y(A) shows more clearly that Y now depends on A.

\(Y(A) = {e^{\frac{{A + 500}}{{80}}}} - 100 \)

b) Household income corresponding to A = 250 CHF per month

This inverse function allows us to determine the income level Y that generates 250 CHF of energy spending per month. The only thing we need to do is replace A in the equation of Y(A) by 250:

\(Y(250) = {e^{\frac{{750}}{{80}}}} - 100 = 11'689.90\) CHF

The solution in words: the monthly income leading to monthly energy spending of 250 CHF is Y = 11,689.90 CHF.

Remark:
If we hadn't solved part a) in the first place, another option would have been to replace A by 250 in the equation of A(Y), and then isolate Y:

\(250 = 80 \cdot \ln (Y + 100) - 500\)

Solving for Y, of course, implies the same calculation as in part a).

c) Energy spending must be 3% of income

There are two different approaches to solve this problem. The second one makes use of linear functions (see the chapter on functions of the preparatory course as a brush up, if necessary).

Approach 1:

"A is 3% of Y" means nothing else but

\(\frac{A}{Y} = \frac{3}{{100}} = 0.03\)

Now replace A by its equation to get

\(\frac{80 \cdot \ln (Y + 100) - 500}{Y} = 0.03\).

Approach 2:

Since energy spending is supposed to be 3% of income, we have a linear function:

\(A(Y) = 0.03Y\)

(to see why, remember that "3 percent of Y" \(= \frac{3}{{100}}\) of \(Y = 0.03 \cdot Y \; \)).

Yet we still have :

\(A(Y) = 80 \cdot \ln (Y + 100) - 500\)

Thus we can equate both terms, i.e. we have the new equation

\(80 \cdot \ln (Y + 100) - 500 = 0.03Y\)

See here for the two graphs.

Solving the equation:

Note that both approaches are equivalent, i.e. they have the same solution set. This is because the equation from approach 1 can be transformed into the equation from approach 2 by multiplying with Y.

To solve any of these equations, we need numeric tools like the one your calculator has. Since both the variable and its own logarithm are part of the equation, we cannot solve it without such a tool.

The solver of our calculator shows us that Y = 6,969.43 CHF.

You might, however, obtain the solution Y = 532.5 CHF. This depends on the starting value of the numerical search.

But: the second solution is not in the domain of A (remember that this was defined to be all Y of at least 1,000 CHF), so we ignore it. The diagram illustrating approach 2 show both solutions.

a) Inverse function

Solve the equation for Y:

\(A = 80 \cdot \ln (Y + 100) - 500\)

\(A+500 = 80 \cdot \ln (Y + 100) \)

\(\frac{{A + 500}}{{80}} = \ln (Y + 100)\)

\({e^{\frac{{A + 500}}{{80}}}} = Y + 100\),

\(Y(A) = {e^{\frac{{A + 500}}{{80}}}} - 100 \)

b) Household income for A = 250 CHF

Replace A by 250 in the above equation:

\(Y(250) = {e^{\frac{{750}}{{80}}}} - 100 = 11'689.90\) CHF

c) Energy spending must be 3%

\(\frac{A}{Y} = \frac{3}{{100}} = 0.03\)

\(A(Y) = 0.03Y\)

\(80 \cdot \ln (Y + 100) - 500 = 0.03Y\)

The solver of the calculator shows that Y = 6'969.43 CHF.
(The second solution Y = 532.5 CHF is not in the domain of the function, so we ignore it.)