<p><span class="math-container">$U(x^\star,y^\star) = a \ln( \frac{am}{p_1}) + b \ln(\frac{bm}{p_2}) $</span></p>
<p><span class="math-container">$= a \ln(a) + a \ln(m) - a \ln(p_1) + b \ln(b) + b \ln(m) - b \ln(p_2) $</span></p>
<p>Since <span class="math-container">$a + b = 1$</span>,</p>
<p><span class="math-container">$= \ln(m) - a \ln(p_1) - b \ln(p_2) + [a \ln(a) + b \ln(b)] $</span></p>
<p><span class="math-container">$= V(p_1,p_2,m) + [a \ln(a) + b \ln(b)]$</span></p>
<p>I got your <span class="math-container">$V$</span> plus some bracket.</p>
<p>So we need the <span class="math-container">$[\text{bracket}] = 0$</span>.</p>
<p><span class="math-container">$a \ln(a) + b \ln(b) = ln(a^a) + ln(b^b) = ln(a^a b^b) = 0$</span></p>
<p>Exponentiating,</p>
<p><span class="math-container">$a^a b^b = 1$</span></p>
<p>Since <span class="math-container">$a+b=1$</span> we can rewrite this condition in terms of a single variable as</p>
<p><span class="math-container">$a^a (1-a)^{1-a} = 1$</span></p>
<p>We now study the graph of <span class="math-container">$f(a) = a^a (1-a)^{1-a}$</span> by using derivatives.</p>
<p>Note the interior of the domain of this function is <span class="math-container">$(0,1)$</span>, with the function being defined limit wise at the endpoints.</p>
<p>To be able to apply the log rule, we take the log of the function</p>
<p><span class="math-container">$\ln(f) = a \ln(a) + (1-a) \ln(1-a)$</span></p>
<p>Differentiating,</p>
<p><span class="math-container">$\frac{f’}{f} = \ln(a) + 1 - \ln(1-a) - 1$</span></p>
<p><span class="math-container">$\frac{f’}{f} = \ln(a) - \ln(1-a)$</span></p>
<p><span class="math-container">$\frac{f’}{f} = \ln(\frac{a}{1-a})$</span></p>
<p><span class="math-container">$f’(a) = a^a (1-a)^{1-a} \ln(\frac{a}{1-a})$</span></p>
<p>The only factor of the derivative that could be 0 at an interior point of the domain is <span class="math-container">$\ln(\frac{a}{1-a})$</span>.</p>
<p>Therefore,</p>
<p><span class="math-container">$f’(a) = 0 \iff \ln(\frac{a}{1-a}) = 0 \iff \frac{a}{1-a} = 1$</span></p>
<p><span class="math-container">$\iff a = 1-a \iff 2a = 1 \iff a = \frac{1}{2}$</span></p>
<p>So we got an extremum of <span class="math-container">$f$</span> at <span class="math-container">$a = \frac{1}{2}$</span>.</p>
<p>By the closed interval theorem, we calculate the values of <span class="math-container">$f$</span> at the limit points to get the range of <span class="math-container">$f$</span>.</p>
<p>Intuitively, the fact that we got a single interior extremum implies the graph of <span class="math-container">$f$</span> is U-shaped.</p>
<p>I’m going to assume you recall from differential calculus that <span class="math-container">$lim_{x \to 0} x^x = 1$</span>.</p>
<ul>
<li>At <span class="math-container">$a = 0$</span></li>
</ul>
<p><span class="math-container">$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 $</span></p>
<ul>
<li>At <span class="math-container">$a = 1$</span></li>
</ul>
<p><span class="math-container">$\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 $</span></p>
<p>At the interior extremum <span class="math-container">$a = \frac{1}{2}$</span></p>
<p><span class="math-container">$f(\frac{1}{2}) = (\frac{1}{2})^\frac{1}{2} \cdot (\frac{1}{2})^\frac{1}{2} = \frac{1}{2}$</span></p>
<p>This implies <span class="math-container">$f$</span> goes from <span class="math-container">$1$</span> down to <span class="math-container">$\frac{1}{2}$</span>, then back up to <span class="math-container">$1$</span>.</p>
<p>Therefore, the condition <span class="math-container">$f(a) = 1$</span> is satisfied exactly at the endpoints <span class="math-container">$a = 0$</span> and <span class="math-container">$a = 1$</span>.</p>
<p>Recalling <span class="math-container">$a+b = 1$</span>, we can conclude that <span class="math-container">$(a,b) = (1,0) \text{ or } (0,1)$</span>.</p>
<p>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 <span class="math-container">$\frac{m}{p_i}$</span> units of it.</p>
<p>Here is the graph of the function <span class="math-container">$f(a) = a^a (1-a)^{1-a}$</span></p>
<p><a href="https://i.sstatic.net/6aO15.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6aO15.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-05-22T17:33:36.073Z
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?
CormJack
CC BY-SA 4.0
2023-05-22T17:34:55.423Z
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.
CormJack
CC BY-SA 4.0
2023-05-22T17:40:10.620Z
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?
CormJack
CC BY-SA 4.0
2023-05-22T17:42:32.033Z
@CormJack Yes, that was what I meant. I used the log rule and the chain rule: $\frac{d}{da} [\ln(f(a))] = \frac{f’(a)}{f(a)}$
Nicolas Torres
CC BY-SA 4.0
2023-05-22T17:45:28.310Z
@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.
Nicolas Torres
CC BY-SA 4.0
2023-05-22T17:26:13.087Z
2023-05-22T17:33:11.100Z
2023-05-22T17:33:11.100Z
Nicolas Torres
<p><strong>Question:</strong></p>
<p>I'm told the following (by an exam mark scheme):</p>
<p>Using <span class="math-container">$a + b =1$</span></p>
<p><span class="math-container">$a[ln(\frac{am}{p_1})] + b[ln(\frac{bm}{p_2})] = ln(m) - aln(p_1) - bln(p_2)$</span></p>
<p>I can't get this to hold without the additional assumption <span class="math-container">$a^ab^b = 1$</span></p>
<p>Because my Log equation becomes:</p>
<ul>
<li><span class="math-container">$ln\frac{m}{p_1^ap_2^b} = ln\frac{m^{a+b}a^ab^b}{p_1^ap_2^b}$</span></li>
</ul>
<p><strong>Context:</strong></p>
<p>We are given</p>
<p><span class="math-container">$U(x_1,x_2) = aln(x_1) + bln(x_2) \\ s.t. \\  x_1p_1 +x_2p_2 = m$</span></p>
<p>And a Value function <span class="math-container">$v(p_1, p_2, m) = ln(m) - aln(p_1) - bln(p_2)$</span></p>
<p>We have to find the condition for <span class="math-container">$a$</span> and <span class="math-container">$b$</span> such that</p>
<p>Given the Marshellian demand functions:</p>
<ul>
<li><span class="math-container">$x_1 = \frac{am}{p_1}$</span></li>
<li><span class="math-container">$x_2 = \frac{bm}{p_2}$</span></li>
</ul>
<p><span class="math-container">$U(x_1,x_2) = V$</span></p>
<p>I.e. <span class="math-container">$aln(\frac{am}{p_1}) + bln(\frac{bm}{p_2}) = ln(m) - aln(p_1) - bln(p_2)$</span></p>
<p>I find that the condition <span class="math-container">$a +b = 1$</span> is not enough?</p>

<mathematical-economics><consumer-theory><optimization><lagrangian>
Logarithmic Utility function Algebra
CormJack
# Logarithmic Utility function Algebra

(Asked by: username_0 on 2023-05-21)

**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?

**Tags:** mathematical-economics, consumer-theory, optimization, lagrangian

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## 1 answers

### Answer by username_1 on 2023-05-22. Score: 1 | Accepted answer

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

[![enter image description here](https://i.sstatic.net/6aO15.png)](https://i.sstatic.net/6aO15.png)

> **Comments:**
> **username_0**: 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?  
> **username_0**: 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.  
> **username_0**: 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?  
> **username_1**: @CormJack Yes, that was what I meant. I used the log rule and the chain rule: $\frac{d}{da} [\ln(f(a))] = \frac{f’(a)}{f(a)}$  
> **username_1**: @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.  

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