I've got the power! (Mathathon Problem 3)

$4 < 5$

$\text{log } 4 < \text{log } 5$

$\dfrac {\text{log } 4}{\text{log } 4} < \dfrac {\text{log } 5}{\text{log } 4}$ (The sign of the inequality doesn't change since log 4 is +ve (>0))

$1 < \text{log }_4 5$

$1 < \dfrac{8^3}{8^3} \text{log }_4 5$

$1 < \text{log }_{4^{8^3}} 5^{8^3}$

$8^3 = (2^3)^3 = 2^9$

$1 < \text{log }_{(2^2)^{(2^9)}} 5^{8^3}$

$1 < \text{log }_{2^{2^10}} 5^{8^3}$

$1 < \dfrac {\text{log } 5^{8^3}}{\text{log }2^{2^{10}}}$

$\text{log } 2^{2^{10}} < \text{log } 5^{8^3}$ ( $\text{log } 2^{2^{10}} > 0)$

$\boxed{2^{2^{10}} < 5^{8^3}}$

$\textcolor{#D61F06}{2^{2^{10}}}$ [?] $\textcolor{#95D3FE}{5^{8^{3}}}$

We all know,

$\textcolor{#D61F06}{ 2*2*2}$ (i.e.) $\textcolor{#D61F06}{2^{3}= 8}$

hence substituting this we get,

$\textcolor{#D61F06}{2^{2^{10}}}$ [ ?] $\textcolor{#95D3FE}{5^{2^{3*3}}}$ [ $\textcolor{#20A900}{since (a^{m})^{n} =a^{m.n}}$ ]

Now, taking log on both sides( $\textcolor{#3D99F6}{why?}$

1.logs make calculations easier by taking the powers of the variables/constants/functions outside and hence prone to cancellation .

2. log rhymes with the word dog (s)( who never misuse their $\textcolor{#CEBB00}{powers}$ .)

$\textcolor{#D61F06}{2^{10} * log_{10} 2}$ [?] $\textcolor{#95D3FE}{2^{9} * log_{10} *5}$

And $\textcolor{#20A900}{2^{10}\text{ is nothing but}} 2 * 2^{9}$

Voilà! we are now just left with,

$\textcolor{#D61F06}{2 *log_{10} 2}$ and $\textcolor{#95D3FE}{log_{10} 5}$

As facts we know that,

$\textcolor{#3D99F6} {log_{10} 2 =0.3}$ and $\textcolor{#3D99F6} {log_{10} 5 =0.69}$

Soooooo which one's greater $2 * 0.3 [or ]0.69$ ?

Pretty obvious right,

$\textcolor{#69047E}{0.6 [<] 0.69}$

Hence, option [b] is correct!

$\textcolor{#D61F06}{2^{2^{10}}}$ [<] $\textcolor{#95D3FE}{5^{8^{3}}}$

Honestly, what I did to solve this question is crunched the number using a calculator and see which had more digits

$2^{2^{10}} = 1.797693134862315907729305190789$ E+308

$5^{8^3} = 7.4583407312002067432909653154629$ E+357

'E' stands for the power of 10.

for example : $1.2 \textbf {E+5} = 120000$

As the base numbers are positive, both numbers after simplifying will be positive, Due to which the number with larger digits will be greater.

Hence $5^{8^3}$ is greater

Okay, so first note that $8^3 = (2^3)^3 = {2^9}$ . Therefore $5^{8^3} = 5^{2⁹}$ . Now we can compare using logarithms. We will assume that $5^{2⁹}>2^{2^{10}}$ . $\begin{aligned} 5^{2⁹}&>2^{2^{10}}\quad\mid \log_2\\ \log_2(5^{2⁹}) &> 2^{10}\\ 2^9\log_2(5) &> 2^{10}\quad\mid :2^{9}\\ \log_2(5) &> 2\\ \log_2(4) = 2 \implies \log_2(5) &> 2 \end{aligned}$ Tadaa, our assumption was correct!

First converting the second exponent into a power of two, we get $8^3=(2^3)^3=2^9$ , so $8^3=2^9$ . Using this, we have $5^{8^{3}}=5^{2^{9}}$ .
Taking the logarithm of both $2^{2^{10}}$ and $5^{2^{9}}$ , we can see that $\log 2^{2^{10}}$ = $2^{10}\times\log2$ and $\log5^{2^{9}}=2^{9}\times\log5$ . Dividing both expressions by $2^9$ , we then have $2\times\log2$ and $\log5$ . Since $2\times\log(2)=\log(2^2)=\log(4)$ , $\log5>\log4$ , so $5^{8^{3}}>2^{2^{10}}$ .

Step 1

There are two ways to compare positive exponents:
(1) Compare base when power is positive and equal.
(2) Compare power when base is equal.
Time to choose.

Step 2

(1)

You tried and tried but you couldn’t easily convert it into the same base. You died of brain explosion.
Return to step 1.

(2)

By calculation, $2^{2^{10}}=2^{1024}.$ $5^{8^3}=5^{512}.$ (1) Continue to convert.
(2) Use logarithms.

Step 3

(1)

Looking carefully at the powers of the two exponents, we find $1024=2\times 512$ .

By one rule of exponents $\color{#D61F06}a^{mn}=(a^m)^n$ , $2^{1024}=2^{2\times 512}=(2^2)^{512}=4^{512}$ .
We have converted comparing $2^{2^{10}}$ and $5^{8^3}$ to comparing $4^{512}$ and $5^{512}$ .
Since $4\lt 5$ ,
$4^{512}\lt 5^{512}$ .
i.e. $\color{#D61F06}2^{2^{10}}\lt 5^{8^3}$ .
Succeed!

(2)

Of course you can choose to use logarithms. I like using $\log_{10}$ , which is usually simplified as $\log$ .
We can compare $\log 2^{1024}$ and $\log 5^{512}$ instead.
By rule $\color{#D61F06} \log a^m =m\log a,$ $\log 2^{1024}=1024\log 2 = 1024\times 0.301029... =308.2547...$
$\log 5^{512}=512\log 5=512\times 0.69897...=357.87264...\color{#D61F06}\gt 308.2547...$
Since $\log 5^{512}$ is greater, $5^{512}$ is greater, i.e.
$\color{#D61F06} 5^{8^3}\text{is greater.}$
Succeeded again!

2^10 = 1024 , 8^3 = 512

We have to compare 2^1024 with 5^512

But 2^1024 = 2^(2*512) = (2^2)^512 = 4^512 < 5^512. Therefore 5^8^3 > 2^2^10

By simplifying exponents, we want to which one out of $2^{1024}$ and $5^{512}$ is greater. But since $2^{1024} = 4^{512}$ , it's clear that $5^{512} > 4^{512} = 2^{1024}$ .

First method:

Let's try to make the powers same.

First we see that $8^3 = (2^3)^3 = 2^9$ and $2^9 = 512$

So $5^{8^3} = 5^{2^9}=5^{512}$

Also, $2^{10} = 1024$

$2^{2^{10}} = 2^{1024} = (2^2)^{512} = 4^{512}$

Since $5 > 4$ we get $5^{512}>4^{512}$ and then $5^{512}=5^{8^3}>4^{512}=2^{2^{10}}$

Or we can say that $\sqrt[512]{5^{512}} = 5^{\frac{512}{512}} = 5^1 = 5 > \sqrt[512]{4^{512}} = 4^{\frac{512}{512}} = 4^1 = 4$ and then $5^{512}=5^{8^3}>4^{512}=2^{2^{10}}$

Second method(Brute Force):

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a = 2**2**10
b = 5**8**3

if (max(a,b) == a):
    print("2**2**10 is greater")
else:
    print("5**8**3 is greater")

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#output
"5**8**3 is greater"

$\begin{aligned} &\large{4 < 5} \ &\textcolor{#3D99F6}{\text{Obviously}} \\ \iff &\large{2^2 < 5} &\textcolor{#3D99F6}{4=2^2} \\ \iff &\large{{2^{2^{10}}} < {5^{2^9}}} \ &\textcolor{#3D99F6}{\text{Raising both sides to } 2^{9}} \\ \small{\textcolor{#D61F06}{\text{Note:Both sides are greater than 1,}}}&\small{\textcolor{#D61F06}{\text{ power is greater than 1, so preserving the inequality.}}} \\ \iff & \large{{2^{2^{10}}} < {5^{8^3}}} \ &\textcolor{#3D99F6}{\text{Answer}} \\ \end{aligned}$

First we need to simplify $8^{3}$ into a power of $2$ , so $8^{3} = (2^{3})^{3} = 2^{9}$ . Since $5 > 4$ , $5^{2^{9}} > 4^{2^9}$ , and $4^{2^9} = (2^2)^{2^9} = 2^{2^{10}}$ . So $\boxed{5^{8^{3}}}$ is greater.

$\blue{\because 5>4 \space and \space\log\space is\space an\space increasing\space function}$ $\log_2(\red{5}^{8^3})>\log_2(\red{4}^{8^3})\space$ $=8^3\log_24=(2^3)^3\log_22^2$ $=2^{3\times 3}\times 2=2^{9+1}$ $=2^{10}=2^{10}\times \red{1}$ $=2^{10}\red{\log_22}=\log_22^{2^{10}}$ $\Rightarrow \log_25^{8^3}>\log_22^{2^{10}}$ $\blue{\because \log\space is\space an\space increasing\space function}$ $\therefore \boxed{5^{8^3}>2^{2^{10}}}$

Note:

• $f(x)$ is an increasing function $\iff (f(x_1)>f(x_2)\iff x_1>x_2)$

First $2^{2^{10}}$

• = $2^{2^9×2}$
• Using $a^{mn}$ = $(a^m)^{n}$
• = $4^{2^9}$
• = $4^{8^3}$

And obviously $4^{8^3}$ < $5^{8^3}$ , therefore $2^{2^{10}}$ < $5^{8^3}$

The way I look at it is that $2^{10 } = 2 * 2^9 = 2 * 8^3$

Therefore $2^{2^{10}} = 4^{8^3}$ and $4^{8^3}$ is clearly less than $5^{8^3}$ QED

2^2^10= $2^{1024}$ and 5^8^3= $5^{512}$ Since $\frac{1024}{512}=2$ and $k \sqrt{m^{n}}$ = $m^\frac{n}{k}$ we, $512 \sqrt{2^{1024}}$ = $2^\frac{1024}{512}$ = $2^{2}$ = $4$ and $512 \sqrt{5^{512}}$ = $5^\frac{512}{512}$ = $5^{1}$ = $5$ . Since $4$ is greater than $5$ , {5^8^3} is bigger.

${2^2}^{10}= 2^{2×2×2×2×2×2×2×2×2×2}= 2^{1024}=2×2×2×2× ....×2×2= 4×4×...×4=4^{512}$

${5^8}^3= {5^{(2×2×2)}}^3= 5^{2×2×2×2×2×2×2×2×2}= 5^{512}$

Thus the comparison between ${2^2}^{10}$ and ${5^8}^3$ is the same as it is between $4^{512}$ and $5^{512}$ , where due to powers being same, using the fact that $a^x<b^x$ if $a<b$ , we conclude $4^{512}<5^{512}$ or ${2^2}^{10}<{5^8}^3$ , thus second is larger .

Most Common Approach -

$2^{10} = 1024 \Rightarrow 2^{2^{10}} = 2^{1024} = 4^{512}$

$8^{3} = 512 \Rightarrow 5^{8^{3}} = 5^{512}$

$5^{512} > 4^{512}$

Thus, the answer is $\Large \boxed{5^{8^{3}}}$

See only one solver (limited time)