# Logarithmic Utility function Algebra

Question:

I'm told the following (by an exam mark scheme):

Using $$a + b =1$$

$$a[ln(\frac{am}{p_1})] + b[ln(\frac{bm}{p_2})] = ln(m) - aln(p_1) - bln(p_2)$$

I can't get this to hold without the additional assumption $$a^ab^b = 1$$

Because my Log equation becomes:

• $$ln\frac{m}{p_1^ap_2^b} = ln\frac{m^{a+b}a^ab^b}{p_1^ap_2^b}$$

Context:

We are given

$$U(x_1,x_2) = aln(x_1) + bln(x_2) \\ s.t. \\ x_1p_1 +x_2p_2 = m$$

And a Value function $$v(p_1, p_2, m) = ln(m) - aln(p_1) - bln(p_2)$$

We have to find the condition for $$a$$ and $$b$$ such that

Given the Marshellian demand functions:

• $$x_1 = \frac{am}{p_1}$$
• $$x_2 = \frac{bm}{p_2}$$

$$U(x_1,x_2) = V$$

I.e. $$aln(\frac{am}{p_1}) + bln(\frac{bm}{p_2}) = ln(m) - aln(p_1) - bln(p_2)$$

I find that the condition $$a +b = 1$$ is not enough?

$$U(x^\star,y^\star) = a \ln( \frac{am}{p_1}) + b \ln(\frac{bm}{p_2})$$

$$= a \ln(a) + a \ln(m) - a \ln(p_1) + b \ln(b) + b \ln(m) - b \ln(p_2)$$

Since $$a + b = 1$$,

$$= \ln(m) - a \ln(p_1) - b \ln(p_2) + [a \ln(a) + b \ln(b)]$$

$$= V(p_1,p_2,m) + [a \ln(a) + b \ln(b)]$$

I got your $$V$$ plus some bracket.

So we need the $$[\text{bracket}] = 0$$.

$$a \ln(a) + b \ln(b) = ln(a^a) + ln(b^b) = ln(a^a b^b) = 0$$

Exponentiating,

$$a^a b^b = 1$$

Since $$a+b=1$$ we can rewrite this condition in terms of a single variable as

$$a^a (1-a)^{1-a} = 1$$

We now study the graph of $$f(a) = a^a (1-a)^{1-a}$$ by using derivatives.

Note the interior of the domain of this function is $$(0,1)$$, with the function being defined limit wise at the endpoints.

To be able to apply the log rule, we take the log of the function

$$\ln(f) = a \ln(a) + (1-a) \ln(1-a)$$

Differentiating,

$$\frac{f’}{f} = \ln(a) + 1 - \ln(1-a) - 1$$

$$\frac{f’}{f} = \ln(a) - \ln(1-a)$$

$$\frac{f’}{f} = \ln(\frac{a}{1-a})$$

$$f’(a) = a^a (1-a)^{1-a} \ln(\frac{a}{1-a})$$

The only factor of the derivative that could be 0 at an interior point of the domain is $$\ln(\frac{a}{1-a})$$.

Therefore,

$$f’(a) = 0 \iff \ln(\frac{a}{1-a}) = 0 \iff \frac{a}{1-a} = 1$$

$$\iff a = 1-a \iff 2a = 1 \iff a = \frac{1}{2}$$

So we got an extremum of $$f$$ at $$a = \frac{1}{2}$$.

By the closed interval theorem, we calculate the values of $$f$$ at the limit points to get the range of $$f$$.

Intuitively, the fact that we got a single interior extremum implies the graph of $$f$$ is U-shaped.

I’m going to assume you recall from differential calculus that $$lim_{x \to 0} x^x = 1$$.

• At $$a = 0$$

$$lim_{a \to 0} a^a (1-a)^{1-a} = \lim_{a \to 0} a^a \cdot \lim_{a \to 0} (1-a)^{1-a} = 1 \cdot 1 = 1$$

• At $$a = 1$$

$$\lim_{a \to 1} a^a (1-a)^{1-a} = \lim_{a \to 1} a^a \cdot \lim_{a \to 1} (1-a)^{1-a} = 1 \cdot \lim_{b \to 0} b^b = 1$$

At the interior extremum $$a = \frac{1}{2}$$

$$f(\frac{1}{2}) = (\frac{1}{2})^\frac{1}{2} \cdot (\frac{1}{2})^\frac{1}{2} = \frac{1}{2}$$

This implies $$f$$ goes from $$1$$ down to $$\frac{1}{2}$$, then back up to $$1$$.

Therefore, the condition $$f(a) = 1$$ is satisfied exactly at the endpoints $$a = 0$$ and $$a = 1$$.

Recalling $$a+b = 1$$, we can conclude that $$(a,b) = (1,0) \text{ or } (0,1)$$.

Intuitively, this means that only one of the goods would yield utility, and the consumer would simply spend all their income on that good, buying $$\frac{m}{p_i}$$ units of it.

Here is the graph of the function $$f(a) = a^a (1-a)^{1-a}$$

• Hi Nicholas, firstly thank you so much for the detail, it's really great and quite fascinating! Secondly my course mark scheme makes zero attempt to demonstrate any of this. they don't even mention the weird extra bracket we both get).. They simply stop at saying $α + β = 1$ . Yet as you have shown a truly accurate answer requires more detail. I'm almost certain (99.999%) that they would have not expected anything like what you have written int he exam? Which then leaves the question how they were so comfortable in simply stating $α + β = 1$ is all we needed? Commented May 22, 2023 at 17:33
• I'm happy to say i got this far $a^a (1-a)^{1-a} = 1$ before checking the mark scheme and seeing they made absolutely no mention to it, and then turning to the wizards here like yourself. Commented May 22, 2023 at 17:34
• Could you kindly just clarify the $\frac{f'}{f}$ differentiation, feel free to say roughly in words what's happening here, do you literally just mean derivative of f with respect to a, divided by f, gives us those sums? If so, what was the inspiration for doing this? Commented May 22, 2023 at 17:40
• @CormJack Yes, that was what I meant. I used the log rule and the chain rule: $\frac{d}{da} [\ln(f(a))] = \frac{f’(a)}{f(a)}$ Commented May 22, 2023 at 17:42
• @CormJack I took $\ln$ to get rid of the $a$’s in the exponents (with $\ln$ properties), as it’s otherwise impossible to differentiate when the variable is in both the base and exponent. Commented May 22, 2023 at 17:45