·2015年4月-6月份信息采用情况的通报

Question

Without using a calculator or a computer can you determine which of these two numbers is bigger: 2.25^3.375 or 3.375^2.25?

Answer 1

They are

actually exactly the same number.

• You need to write each base as a power of the same number. $ ??2.25=\frac94=\Bigl(\frac32\Bigr)^{\!2} ??\qquad ??3.375=\frac{27}{8}=\Bigl(\frac32\Bigr)^{\!3} ??$

• Then Substitute those expressions:

$$ \begin{aligned} 2.25^{3.375} &=\Bigl(\bigl(\tfrac32\bigr)^{2}\Bigr)^{\;(\tfrac32)^{3}} =\Bigl(\tfrac32\Bigr)^{\,2\;(\tfrac32)^{3}}\\ 3.375^{2.25} &=\Bigl(\bigl(\tfrac32\bigr)^{3}\Bigr)^{\;(\tfrac32)^{2}} =\Bigl(\tfrac32\Bigr)^{\,3\;(\tfrac32)^{2}} \end{aligned} $$

• Lastly compare the exponents of $\tfrac32$:

$$ 2\bigl(\tfrac32\bigr)^{3}=2\cdot\frac{27}{8}=\frac{27}{4}=6.75 \qquad 3\bigl(\tfrac32\bigr)^{2}=3\cdot\frac{9}{4}=\frac{27}{4}=6.75 $$

Since both powers reduce to $\bigl(\tfrac32\bigr)^{6.75}$, the two original numbers

are equal.

Equivalently,

raising each to the 4th power gives $\frac{3^{27}}{2^{27}}$ in both cases.

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Additional Math

Key facts about $a^{\,b}=b^{\,a}$

step	pick $n$	choose $a_n=(1+\tfrac1n)^{n}$	and $b_n=(1+\tfrac1n)^{n+1}$	result
1	$n=1$	$a_1=2$	$b_1=4$	$2^{4}=4^{2}=16$
2	$n=2$	$a_2=2.25$	$b_2=3.375$	$2.25^{3.375}=3.375^{2.25}$
?	any $n\ge1$	$a_n$ lies below e	$b_n=a_n(1+1/n)$ lies above e	still $a_n^{\,b_n}=b_n^{\,a_n}$

Why it works

1. From $a^{\,b}=b^{\,a}$ we get $\dfrac{\ln a}{a}=\dfrac{\ln b}{b}$. Picking consecutive points on each side of the maximum of $f(x)=\dfrac{\ln x}{x}$ (which occurs at $x=e$) guarantees equality.

2. The family above arises by taking those two points to be in the fixed ratio $ \dfrac{b_n}{a_n}=1+\dfrac1n$.

3. Limit as $n\to\infty$:

   $$ a_n=(1+\tfrac1n)^{n}\;\longrightarrow\;e \qquad b_n=a_n\Bigl(1+\tfrac1n\Bigr)\;\longrightarrow\;e $$

   because $(1+1/n)^{n}$ is the classic defining sequence for $e$ and the extra factor $(1+1/n)$ tends to 1. Thus both sequences “pinch’’ toward Euler’s number while preserving the identity $a_n^{\,b_n}=b_n^{\,a_n}$

Answer 2

This is the second equality in a sequence going back to Euler (the first one is the familiar $2^4=4^2$).

Let

$x=\left(\dfrac{n+1}{n}\right)^{n}$

$y=\left(\dfrac{n+1}{n}\right)^{n+1}$

Then

$x^y=\left(\dfrac{n+1}{n}\right)^{n×\left(\dfrac{n+1}{n}\right)^{n+1}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{(n+1)^{n+1}}{n^n}\right)}$

and

$y^x=\left(\dfrac{n+1}{n}\right)^{(n+1)×\left(\dfrac{n+1}{n}\right)^{n}}$ $=\left(\dfrac{n+1}{n}\right)^{\left(\dfrac{(n+1)^{n+1}}{n^n}\right)}$

Thereby $x^y=y^x$. Here

$x=2.25=9/4=(3/2)^2, y=3.375=27/8=(3/2)^3$

which matches the Euler form with $n=2$. The next member of the sequence, with $n=3$, would be

$(64/27)^{256/81}=(256/81)^{64/27}$

Euler also showed that the formula above contains all the positive rational solutions for $x^y=y^x$ apart from $y=x$.

The values of $ x^y=y^x$ with this formula decrease with increasing $n$. Thus $16=2^4=4^2$ is greater than the limiting value of $e^e(\approx 15.154)$, with $2.25^{3.375}$ and other values in the sequence falling in-between.