Hey guys. I was wondering what would be the quickest way to solve.

Problem:

A: 6^100

B: 100^50

Answer B

Hey @Ballam,

I’m no expert, but if I had to solve it, this how I’d go about it:

A: 6^{100}

B: 100^{50}

B can also be written as 10^{2^50}. Further simplifying the exponents we get; 10^{100} (because exponents over exponents get multiplied.)

So now it becomes an obvious choice while comparing 6^{100} and 10^{100}. No need to calculate anything.

hope this helps.

Jeet

Intuitively, if the bases of exponent numbers have such a big gap, in this case, 100 is over 10 times larger than 6, and exponents only have a little gap, you can asume the number with larger base is bigger. However, to ensure the accuracy, better to manipulate with it a little bit.

6^(2*50)

36^50<100^50

We see, for example:

6^{100} vs. 100^{50}

I have a feeling that prime factorization comes into play, but I don’t even know where to start. How does one generally approach such problems?

Edit: LaTeX

I just figured that 100^50 would be 1 with 100 zeros behind it. 10^100 would also be 1 with 100 zeros. But we are asked to compare 6^100. Since 6 is less than 10, then 6^100 would certainly be less than 10^100 therefore 100^50 is greater

Hello friends.

I need some help with this question.

Quant A

6 to the power of 100

Quant B

100 to the power of 50

which one is larger?

I don’t know what is the procedure to solve it

Help

try to manipulate Qty B (Hint: What is the square root of 100?)