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: 6100
B: 10050
B can also be written as 102^50. Further simplifying the exponents we get; 10100 (because exponents over exponents get multiplied.)
So now it becomes an obvious choice while comparing 6100 and 10100. 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?)