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For example: $3^\sqrt5$ versus $5^\sqrt3$

I tried to write numbers as this:

$3^{5^{\frac{1}{2}}}$ and then as $3^{\frac{1}{2}^5}$

But this method gives the wrong answer because $a^{(b^c)} \ne a^{bc}$

Ben Millwood
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Alex
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  • Remember to bracket things to eliminate ambiguity and to avoid confusing yourself. $3^{(5^{1/2})}$. Also, it's not true that $a^{(b^c)} = a^{bc}$ if that's where you went wrong. – wlad Jun 11 '15 at 18:22
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    You're comparing $3^{5^{1/2}}$ and $5^{3^{1/2}}$ – Akiva Weinberger Jun 11 '15 at 18:24
  • One way is to use continued fraction approximations to the sqrts, e.g. see the post linked at the end of this answer,, which compares $,7^\sqrt 8$ vs $8^\sqrt 7$. – Bill Dubuque Jun 11 '15 at 18:41
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    Seems like a hard problem in general… for example, $2^{\sqrt{79}}<79^{\sqrt2}$, but $2^{\sqrt{80}}>80^{\sqrt2}$. $5^{\sqrt{11}}<11^{\sqrt5}$, but $5^{\sqrt{12}}>12^{\sqrt5}$. $6^{\sqrt9}<9^{\sqrt6}$, but $6^{\sqrt{10}}>10^{\sqrt6}$. – Akiva Weinberger Jun 11 '15 at 18:42

6 Answers6

7

Doing a problem like this takes a combination of knowledge about rules of exponentiation and skill at estimation.

First, raise both $3^{\sqrt 5}$ and $5^{\sqrt 3}$ to the power $\sqrt 5$, which gives the numbers $3^5=243$ and $5^{\sqrt{15}}$ using the rule $(a^b)^c = a^{bc}$. This has the effect of removing one of the square roots from an exponent: now we just need to estimate $5^{\sqrt 15}$. And $\sqrt{15}>3.5$, so $$ 5^{\sqrt{15}}>5^{3.5}=5^3\sqrt 5 = 125\sqrt 5. $$ Since $\sqrt 5>2$, we have that $5^{\sqrt{15}}>125\cdot2=250 > 243 = 3^5$. So $$ \fbox{$5^{\sqrt 3} > 3^{\sqrt 5}$}. $$

user134824
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$$5^\sqrt{3}>5^{5/3}=\sqrt[3]{3125}>\sqrt[3]{2187}=3^{7/3}>3^{\sqrt{5}}$$
Another option is to take roots, and compare $3^{1/\sqrt{3}}$ with $5^{1/\sqrt{5}}$. This function increases for $1<x<7.39$, then decreases.

Empy2
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    So the first method is to aproximate the square root as a fraction but how do I come up with that fraction? – Alex Jun 11 '15 at 19:02
  • In reality, check with your calculator $\sqrt{5}=2.236<2.333$, but to prove that, $45<49\Rightarrow3\sqrt{5}<7$ – Empy2 Jun 12 '15 at 00:06
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    > This function increases which function? I've only found a single $x$ in the whole answer, and before this phrase there're only numbers, not functions, so I'm not sure which function you're referring to in the phrase I quoted. – Ruslan Jun 12 '15 at 05:21
  • Sorry, $x^{1/\sqrt{x}}$ – Empy2 Jun 12 '15 at 09:14
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user134824's approach, but with the $\sqrt{3}$ instead, is a bit nicer:

$$ \begin{align} 3^\sqrt{5} &\text{ vs. } 5^\sqrt{3}\\ (3^\sqrt{5})^\sqrt{3} &\text{ vs. } (5^\sqrt{3})^\sqrt{3}\\ 3^{\sqrt{5}\sqrt{3}} &\text{ vs. } 5^{\sqrt{3}\sqrt{3}}\\ 3^\sqrt{15} &\text{ vs. } 5^3\\ \end{align} $$

Now notice that since $\sqrt{15} < 4$, $3^\sqrt{15} < 3^4$ (via monotonicity of $3^x$), and since $3^4 < 5^3$ ($81 < 125$), we can say $3^\sqrt{15}<5^3$ (via transitivity), and subsequently $3^\sqrt{5} < 5^\sqrt{3}$ (via monotonicity of $x^\dfrac{1}{\sqrt{3}}, x \ge 0$).

YawarRaza7349
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Consider

$f(x)=x^{\sqrt{5}}-5^{\sqrt{x}}$

$f(\sqrt{5})=0 $

$f'(x)=(\sqrt{5}-1)x-\frac {log(5)}{2 \sqrt{x}}e^{\sqrt{x}log5}$

so $ f(x) < 0$ for $x$ less than $\sqrt{5}$ so we have $$ 5^{\sqrt{3}} > 3^{\sqrt{5}}$$

RowanS
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$$3^{\sqrt 5} \lt 5^{\sqrt3}$$ $$\sqrt 5 \cdot \ln(3) \lt \sqrt3 \cdot \ln(5)$$ $$\sqrt{5 \over 3} \cdot \log_5 (3) \lt \sqrt{5 \over 3} \cdot {3 \over 4} \lt 1$$ because $\log_a(b) \lt 1$ for $a \gt b$, $\ln(3) \lt {3 \over 2}$, and $\ln(5) \lt 2$

Zach466920
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Hint: Consider the function $$f(x)={\ln x\over \sqrt x}$$ and see where it is increasing or decreasing.

Fermat
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  • The critical point is $x=e^2$ and fortunately $3$ and $5$ are on the same side. This approach would not work for example with $4$ and $16$. – Henry Jun 11 '15 at 20:55
  • Yes. You are right. I examined the function before posting this answer. @Henry – Fermat Jun 11 '15 at 21:59