<p>You need to find an exponent such that the two utility functions represent the same preferences, not such that they are not equal.</p>
<p>These two functions are clearly not the same.</p>
<p>Recall, however, that utility functions represent unique preference relations only up to monotone increasing transformations, meaning that for any utility function:</p>
<p><span class="math-container">$$U(\tilde{x})$$</span></p>
<p>We can also represent the preferences with any function:</p>
<p><span class="math-container">$$\overline{U}(\tilde{x}) = g(U(\tilde{x}))$$</span></p>
<p>Where <span class="math-container">$g(.)$</span> is increasing.</p>
<p>You need, therefore, to find a monotonic transformation of the first utility function that yields a utility function of the form of the second utility function. You're correct that the transformation for this is to take logarthims, and you don't apply it to both functions (you apply <span class="math-container">$g()$</span> only to the first function).</p>
<p><span class="math-container">$$log(u^A) = log(x_1 x_2^\alpha) = log(x_1) + \alpha log(x_2)$$</span></p>
<p>This is clearly not quite of the form <span class="math-container">$u^B$</span>, but it's close. You then need to make an additional transformation, dividing through by <span class="math-container">$(1+\alpha)$</span>, which is also a monotonic transformation so long as <span class="math-container">$\alpha &gt; -1$</span>:</p>
<p><span class="math-container">$$\dfrac{log(u^A)}{1 + \alpha} = \frac{1}{1+\alpha}log(x_1) + \frac{\alpha}{1+\alpha} log(x_2)$$</span>.</p>
<p>Finally, then, you can set <span class="math-container">$\frac{1}{1 + \alpha} = \frac{1}{3} \implies \alpha = 2$</span>.</p>
<hr />
<p>Consequently, your first solution is correct. The reason your latter method doesn't work is because the utility functions aren't <em>equal</em>, they just represent the same preferences.</p>

CC BY-SA 4.0
2024-05-05T07:17:42.417Z
2024-05-05T07:17:42.417Z
matthewoulton
<blockquote>
<p>However, when I set the equations to be equal</p>
</blockquote>
<p>These are not equations, they are functions. You can quite quickly validate that the two functions are not the same function, they are not 'equal', by checking their values at <span class="math-container">$(1,1)$</span>.</p>
<p>You also write</p>
<blockquote>
<p><span class="math-container">$$ \ln(x_1^{1/3} \cdot x_2^{2/3})=  x_1(x_2)^\alpha$$</span>
<span class="math-container">$$ \ln(x_1^{1/3} \cdot x_2^{2/3})=  \ln(x_1)+\alpha \cdot \ln(x_2)$$</span></p>
</blockquote>
<p>but these two equations are not consistent; you only took the logarithm of the right hand side of the first equation.</p>
<p>You also write</p>
<blockquote>
<p><span class="math-container">$$x_1^{1/3} \cdot x_2^{2/3} = x_1 \cdot x_2^{\alpha} $$</span>
<span class="math-container">$$x_1^{1/3} \cdot x_2^{2/3} = x_1^{1/3} \cdot x_2^{\alpha} $$</span></p>
</blockquote>
<p>but again, these two equations have little to do with each other, they are not derived from one another.</p>
<p>I recommend sticking to methods that you have discussed in class or going to the office hours of your lecturer to talk through the details of mathematical operations depending on whether you want to pass or get a better understanding.</p>

CC BY-SA 4.0
2024-05-05T06:19:53.570Z
2024-05-05T06:19:53.570Z
Giskard
<p>Two consumers, A and B have the following utility functions
<span class="math-container">$$ u^{A}(x_1,x_2)=x_1(x_2)^\alpha$$</span>
<span class="math-container">$$u^{B}(x_1,x_2)=\frac{1}{3}\ln(x_1)+\frac{2}{3}\ln(x_2)  $$</span>
I have to find <span class="math-container">$\alpha$</span> such that the utility functions represent identical preferences. When I transform <span class="math-container">$u^{A}$</span>'s utility function with <span class="math-container">$f(x)=\frac{1}{3}\ln(x)$</span> I get that
<span class="math-container">$$(f ∘ u^{A})(x_1,x_2)=\frac{1}{3}\ln(x_1)+ \frac{\alpha}{3}\ln(x_2)$$</span>
Which means that when <span class="math-container">$\alpha =2$</span>, the preferences are identical. However, when I set the equations to be equal
<span class="math-container">$$\frac{1}{3}\ln(x_1)+\frac{2}{3}\ln(x_2)=x_1(x_2)^\alpha$$</span>
<span class="math-container">$$ \ln(x_1^{1/3}+\ln(x_2^{2/3})=  x_1(x_2)^\alpha$$</span>
<span class="math-container">$$ \ln(x_1^{1/3} \cdot x_2^{2/3})=  x_1(x_2)^\alpha$$</span>
<span class="math-container">$$ \ln(x_1^{1/3} \cdot x_2^{2/3})=  \ln(x_1)+\alpha \cdot \ln(x_2)$$</span>
For logarithms to be equal, I set their arguments to be equal:
<span class="math-container">$$x_1^{1/3} \cdot x_2^{2/3} = x_1 \cdot x_2^{\alpha} $$</span>
<span class="math-container">$$x_1^{1/3} \cdot x_2^{2/3} = x_1^{1/3} \cdot x_2^{\alpha} $$</span>
Now the bases are equal, and for the exponents to be equal <span class="math-container">$\alpha$</span> is equal to 2/3.</p>
<p>So my question is, which of the ways is the correct one?</p>

<microeconomics><mathematical-economics>
Positive monotonic transformation and identical preferences
Jonathan
# Positive monotonic transformation and identical preferences

(Asked by: username_0 on 2024-05-05)

Two consumers, A and B have the following utility functions
$$ u^{A}(x\_1,x\_2)=x\_1(x\_2)^\alpha$$
$$u^{B}(x\_1,x\_2)=\frac{1}{3}\ln(x\_1)+\frac{2}{3}\ln(x\_2) $$
I have to find $\alpha$ such that the utility functions represent identical preferences. When I transform $u^{A}$'s utility function with $f(x)=\frac{1}{3}\ln(x)$ I get that
$$(f ∘ u^{A})(x\_1,x\_2)=\frac{1}{3}\ln(x\_1)+ \frac{\alpha}{3}\ln(x\_2)$$
Which means that when $\alpha =2$, the preferences are identical. However, when I set the equations to be equal
$$\frac{1}{3}\ln(x\_1)+\frac{2}{3}\ln(x\_2)=x\_1(x\_2)^\alpha$$
$$ \ln(x\_1^{1/3}+\ln(x\_2^{2/3})= x\_1(x\_2)^\alpha$$
$$ \ln(x\_1^{1/3} \cdot x\_2^{2/3})= x\_1(x\_2)^\alpha$$
$$ \ln(x\_1^{1/3} \cdot x\_2^{2/3})= \ln(x\_1)+\alpha \cdot \ln(x\_2)$$
For logarithms to be equal, I set their arguments to be equal:
$$x\_1^{1/3} \cdot x\_2^{2/3} = x\_1 \cdot x\_2^{\alpha} $$
$$x\_1^{1/3} \cdot x\_2^{2/3} = x\_1^{1/3} \cdot x\_2^{\alpha} $$
Now the bases are equal, and for the exponents to be equal $\alpha$ is equal to 2/3.

So my question is, which of the ways is the correct one?

**Tags:** microeconomics, mathematical-economics


---

## 2 answers

### Answer by username_1 on 2024-05-05. Score: 0 | Accepted answer

You need to find an exponent such that the two utility functions represent the same preferences, not such that they are not equal.

These two functions are clearly not the same.

Recall, however, that utility functions represent unique preference relations only up to monotone increasing transformations, meaning that for any utility function:

$$U(\tilde{x})$$

We can also represent the preferences with any function:

$$\overline{U}(\tilde{x}) = g(U(\tilde{x}))$$

Where $g(.)$ is increasing.

You need, therefore, to find a monotonic transformation of the first utility function that yields a utility function of the form of the second utility function. You're correct that the transformation for this is to take logarthims, and you don't apply it to both functions (you apply $g()$ only to the first function).

$$log(u^A) = log(x\_1 x\_2^\alpha) = log(x\_1) + \alpha log(x\_2)$$

This is clearly not quite of the form $u^B$, but it's close. You then need to make an additional transformation, dividing through by $(1+\alpha)$, which is also a monotonic transformation so long as $\alpha > -1$:

$$\dfrac{log(u^A)}{1 + \alpha} = \frac{1}{1+\alpha}log(x\_1) + \frac{\alpha}{1+\alpha} log(x\_2)$$.

Finally, then, you can set $\frac{1}{1 + \alpha} = \frac{1}{3} \implies \alpha = 2$.

---

Consequently, your first solution is correct. The reason your latter method doesn't work is because the utility functions aren't *equal*, they just represent the same preferences.


---

### Answer by username_2 on 2024-05-05. Score: 0

> However, when I set the equations to be equal

These are not equations, they are functions. You can quite quickly validate that the two functions are not the same function, they are not 'equal', by checking their values at $(1,1)$.

You also write

> $$ \ln(x\_1^{1/3} \cdot x\_2^{2/3})= x\_1(x\_2)^\alpha$$
> $$ \ln(x\_1^{1/3} \cdot x\_2^{2/3})= \ln(x\_1)+\alpha \cdot \ln(x\_2)$$

but these two equations are not consistent; you only took the logarithm of the right hand side of the first equation.

You also write

> $$x\_1^{1/3} \cdot x\_2^{2/3} = x\_1 \cdot x\_2^{\alpha} $$
> $$x\_1^{1/3} \cdot x\_2^{2/3} = x\_1^{1/3} \cdot x\_2^{\alpha} $$

but again, these two equations have little to do with each other, they are not derived from one another.

I recommend sticking to methods that you have discussed in class or going to the office hours of your lecturer to talk through the details of mathematical operations depending on whether you want to pass or get a better understanding.


---

