Calvin loves comparing!

Algebra Level 1

$\Large 2^{500} \quad \quad 3^{300} \quad \quad 5^{200} \quad \quad 4^{100}$

Which of these numbers is the largest?

2^{500}

3^{300}

5^{200}

4^{100}

21 solutions

Abhinandan Padhi
Jun 19, 2015

Just convert all the numbers to the form: n^100. The number having n as the largest is the biggest. ie. 2^500=32^100; 3^300=27^100; 5^200=25^100; 4^100 is already to the 100th power. Hence, 32^100 is largest.

Moderator note:

Simple standard approach.

Bonus question : Can you arrange the numbers $2^{1/500}, 3^{1/300},5^{1/200}, 4^{1/100}$ in ascending order?

For the Challenge Master Note:

NOTE: Let "a^1/n" stand for (a)^(1/n). It is NOT EQUAL TO (a^1)/n. Eg. 2^1/500 represents (2)^(1/500)=2^(0.002). [This is done simply for convenience].

NOW:

2^1/500= (2^1/5)^1/100

3^1/300=(3^1/3)^1/100

5^1/200=(5^1/2)^1/100

4^1/100=(2^2)^1/100

We know that the larger the number n, the larger n to any power will be, provided n>1. So now the question is to compare the numbers inside the brackets: 2^1/5, 3^1/3, 5^1/2, and 2^2. Obviously, 2^2=4 is the largest among the 4.

Next, we compare the other 3 numbers. We do the this in a similar way:

2^1/5= (2^6)^1/30

3^1/3=(3^10)^1/30

5^1/2=(5^15)^1/30

Since all the numbers in the brackets are raised to the same power (1/30), we can easily say that 2^6 < 3^10 < 5^15.

ANSWER: Hence, from smallest we can arrange the numbers as 2^1/500, 3^1/300, 5^1/200, and 4^1/100.

We can write this as: 2^1/500 < 3^1/300 < 5^1/200 < 4^1/100.

Abhinandan Padhi - 5 years, 11 months ago

How we can find the numbers like in this problem he took 1/100 then 1/30 out of all the numbers, I know how to guess but is there any appropriate process to recognise this ?

Syed Baqir - 5 years, 11 months ago

You should use LaTeX.

Shashank Rammoorthy - 5 years, 11 months ago

What is LaTeX?

Souvik Ghosh - 5 years, 7 months ago

@Souvik Ghosh – https://en.wikipedia.org/wiki/LaTeX

Shashank Rammoorthy - 5 years, 7 months ago

Shouldn't it be 32*2^100? And likewise for the others...

A Former Brilliant Member - 5 years, 11 months ago

I think he had it right; for example, if we just look at 2^500, we can then call that (2^5)^100, which we can see is equal to 32^100.

I always like to imagine the numbers this way when I'm working with exponents: (2x2x2x2x2)x(2x2x2x2x2)x...x(2x2x2x2x2) <-- 100th time, which is equal to 2x2x...x2 <-- 500th time.

Nathan Sproul - 5 years, 11 months ago

32*2^100 = 2^5 * 2^100 = 2^105

Davy Ker - 5 years, 11 months ago

((2^ 1 divided by 500) 1 over 500 is 0.002 there for now the question reads at 2 by the power of 0.002. = 1.00 )) now as for the challenge it is incorrect as to it's correctness. Here's why 2 by the powers of 500 is 2 times 2 (500) times, that sound answerable but it's a long and unanswerable process (Your answer is basically a mathematical Error) so the only mathematically calculable answer is 4 by the power of 100 (4^500) but this is cutting it close to the end of calculable numbers 1.6....E.... is all I am gonna bother writing it out as. Because any number that is unobtainable is considered infinite it is given 0 as it's value, as it's unobservable. (0 is an infinite number of unobservable things) therefore since the only equation that has an ability to be observed is 4 by the power of 100 it is the largest observable number. This could mean that all the numbers are correct. But essentially trying to use simple math to calculate an unobservable item is not gonna work. There for to actually find the answer you want you can't use simply 2 by the power of 500, you have to use a larger mathematical process to get the actual final goal. 2 x 2, 500 times how ever does not lead to what it may look like. Using Trig or Algebra to simplify the answer would be far faster and far more efficient the power fractions. meaning this question is flawed.

Cori Ravenheart - 5 years, 11 months ago

this isn't to say the answer is wrong, it is however to say that the form is wrong. Since the numbers can not easily be observed in their current formula (exponents, fractions, or division) the formula has to be recreated into an observable formula. One that today's lack of coffee has been guessing at. but simply put the larger the answer is gonna be, the larger the formula that has to be used. This applies to all unobservable things. One reason our lack of understanding on how the universe came to be is we lack the realization that it didn't. The universe's full scale is largely unobservable. We only see a very unobservable small percent of how many galaxies actually are forming or have formed in this universe. The universe has 0 borders. it has no ups downs, or horizons and thus it is incalculable in it's total age and total size. We can how ever make observations and mathematical formulas to create an answer to how old and how large the parts we can see are. and to make that we need to use large math.

Cori Ravenheart - 5 years, 11 months ago

Take the LCM of 500, 300, 200 & 100, in this case 3000. Make all exponents equal to 3000 and raise the bases to respective powers. We get 2^(6/3000), 3^(10/3000), 5^(15/3000) & 4^(30/3000). Then compare 2^6, 3^10, 5^15 & 4^30. 4^30 is our winner.

I'Still LoveYou - 5 years, 6 months ago

They're already in ascending order!

Davy Ker - 5 years, 9 months ago

Challenge master note: $({2^6})^{1/3000}, ({3^{10}})^{1/3000}, ({5^{15}})^{1/3000}, ({4^{30}})^{1/3000}$

So we are essentially ordering what comes inside each bracket. We can write $4^{30}$ as $8^{15}$ to make it really easy to see that by the nature of exponents:

$2^6 < 3^{10} < 5^{15} < 8^{15}$

and hence they were already in ascending order!

Davy Ker - 5 years, 5 months ago

(log 4)/ 100 > log 5/200 > log 3/300 > log 2/500

Gaurav Kakked - 5 years, 11 months ago

Neil Yabut
Jun 21, 2015

since they're all positive, if you take the 100th root of all, their relative order of size should still be the same.

hence, it'll boil down to putting 2^5 = 32, 3^3 = 9, 5^2 = 25, and 4^1 = 4 in order.

Moderator note:

Exactly! Good job!

Error in above statement 3^3 = 27 and not 9

Guiseppi Butel - 5 years, 11 months ago

What I did was take $\ln$ of all the numbers which gives ,

$500\ln 2 , 300\ln 3 , 200\ln 5 , 100\ln 4$

then divided all of them by 100 which gives ,

$5\ln 2 , 3\ln 3 , 2\ln 5 , \ln 4$

which is ,

$\ln 2^{5} , \ln 3^{3} , \ln 5^{2} , \ln 4$

We know that,

$e^{\log x} = x$

so taking exponential we get,

$2^{5} , 3^{3} , 5^{2} , 4$

which is,

$32 , 27 , 25 , 4$

We can obviously see that 32 is the larger value.So $2^{500}$ is the largest number out of the list

Athiyaman Nallathambi - 5 years, 10 months ago

It is same as above solutions , you are just making it complicated !!

Syed Baqir - 5 years, 9 months ago

I am aware that it is the same as all the above solutions.That's why I didn't post its as a solution and yeah it sure as hell looks complicated but when I was solving the question , it took no more than 20 seconds.

Athiyaman Nallathambi - 5 years, 9 months ago

Dolly Mouni
Jun 26, 2015

simply compare values of 2^5,4^1,5^2,3^3

Reza Lutfi
Jun 26, 2015

easily 2^5=32 3^3=27 5^2=25 4^1=4

Sarat Sahoo
Jun 21, 2015

as simple as that .logic is to know which is big among something ,bring them to a common attribute or measure them in a common way .thats it.here the common thing is make everything raise to the power of 100/

Moderator note:

Yes. Can we solve this via logarithms too?

For Challenge Master note:

We can solve via logarithms.

First take $\ln$ of all the numbers which gives ,

$500\ln 2 , 300\ln 3 , 200\ln 5 , 100\ln 4$

then divide all of them by 100 which gives ,

$5\ln 2 , 3\ln 3 , 2\ln 5 , \ln 4$

which is ,

$\ln 2^{5} , \ln 3^{3} , \ln 5^{2} , \ln 4$

We know that,

$e^{\log x} = x$

so taking exponential we get,

$2^{5} , 3^{3} , 5^{2} , 4$

which is,

$32 , 27 , 25 , 4$

We can obviously see that 32 is the larger value.So $2^{500}$ is the largest number out of the list

Athiyaman Nallathambi - 5 years, 10 months ago

Monisha Nair
Nov 20, 2015

Logarithm being a proportional function, one may take the logarithm of each to compare such numbers.

Iqra Fatimah
Oct 26, 2015

Number having n as a largest is the biggest number.

Sadasiva Panicker
Oct 11, 2015

2^5=32, 3^3=27. 5^2=25 &4^1=4 So 2^500 is the largest.

Wolf 454
Sep 2, 2015

2^500 is a lot!!!Mord gigantic then all others

Athiyaman Nallathambi
Jul 29, 2015

We can find the larger number by using logarithms. First we take $\ln$ of all the numbers which gives ,

$500\ln 2 , 300\ln 3 , 200\ln 5 , 100\ln 4$

then divide all of them by 100 which gives ,

$5\ln 2 , 3\ln 3 , 2\ln 5 , \ln 4$

which is ,

$\ln 2^{5} , \ln 3^{3} , \ln 5^{2} , \ln 4$

We know that,

$e^{\log x} = x$

so taking exponential we get,

$2^{5} , 3^{3} , 5^{2} , 4$

which is,

$32 , 27 , 25 , 4$

We can obviously see that 32 is the larger value.So $2^{500}$ is the largest number out of the list

Upkar Singh
Jul 5, 2015

Converting to n x 100^m and then compare

Sarfaraz Muhammad
Jul 2, 2015

2^500 is the correct answer

Himanshu Ojha
Jul 1, 2015

I just took the lcm of pwers acnd got the answer

Neelesh I
Jun 30, 2015

(2^5)^100 > (3^3)^100 > (5^2)^100 > (4)^100

Shikhar Shukla
Jun 30, 2015

Assume a variable for each and take log base 10 both sides Log(var)=some no. Since for base 10 log graph is continously inc.ng So var with largest no. Will be greatest among all

Pranav Jagadeesan
Jun 28, 2015

Pranav

AUSTRALIA

For the Challenge Master Note: NOTE: Let "a^1/n" stand for (a)^(1/n). It is NOT EQUAL TO (a^1)/n. Eg. 2^1/500 represents (2)^(1/500)=2^(0.002). [This is done simply for convenience]. NOW: 2^1/500= (2^1/5)^1/100 3^1/300=(3^1/3)^1/100 5^1/200=(5^1/2)^1/100 4^1/100=(2^2)^1/100 We know that the larger the number n, the larger n to any power will be, provided n>1. So now the question is to compare the numbers inside the brackets: 2^1/5, 3^1/3, 5^1/2, and 2^2. Obviously, 2^2=4 is the largest among the 4. Next, we compare the other 3 numbers. We do the this in a similar way: 2^1/5= (2^6)^1/30 3^1/3=(3^10)^1/30 5^1/2=(5^15)^1/30 Since all the numbers in the brackets are raised to the same power (1/30), we can easily say that 2^6 < 3^10 < 5^15. ANSWER: Hence, from smallest we can arrange the numbers as 2^1/500, 3^1/300, 5^1/200, and 4^1/100. We can write this as: 2^1/500 < 3^1/300 < 5^1/200 < 4^1/100.

Clayton Knittel
Jun 27, 2015

Raise each number to the 1/100th power. Then compare the results.

Mohammad Mahmoud
Jun 25, 2015

2^500 = 32^100, 5^200 = 25^100, 3^300 = 27^100, and we have 4 ^100

Anil Kumar
Jun 21, 2015

Apply log to all...better to apply log 2 than log10 since estimating log2 values for this problem is easy...log5 base2 = 2.2(approx),log3base2= 1.6(approx)...so you can do it without calculator

Moderator note:

Of course you need to have the log tables to solve this.

Gaurav Kakked
Jun 21, 2015

Applying Log on both sides ... 500log 2 > 300log3 > 200 log 5 > 100 log 4...

Moderator note:

How are we suppose to show that "500log 2 > 300log3 > 200 log 5 > 100 log 4" is true without log tables?

AwaIs Ali
Jun 21, 2015

2^5 * 100, 3^3 * 100, 5^2 * 100, 4^1 * 100

now, exclude all 100's from the powers.

And the answer is : 32, 9, 25, 4 So, 2^500 is larger.

and also 2 is the smallest base through which we got the largest number. so we don't need to go on and compare with others.

Moderator note:

Why can we "exclude all 100's from the powers."?

Calvin loves comparing!

21 solutions

Moderator note:

Moderator note:

Moderator note:

Moderator note:

Moderator note:

Moderator note:

0 pending reports