Question

Using only paper and pencil (no calculator or logarithmic tables), determine which of the following expressions has a greater value: 10^1/10 or 3^1/3.

Answer 1

The first one to get it gets a medal

Answer 2

@iGreen

Answer 3

no, I know the answer, I just wanted to see if anyone could do it XD

Answer 4

Lol, I have no idea how to do this stuff.

Answer 5

oh XD

Answer 6

So, I'm assuming you know @phi?

Answer 7

oops, it should be
x^10 = 10

y^3 = 3,

but working with those should tell us the answer

Answer 8

notice (y^3) squared is 9
i.e. y^6 = 9
x^10 = 10
9 and 10 are close to the same value
but it takes y times itself only 6 times to get to 9
so y will be bigger than x
i.e.
3^(1/3) > 10^(1/10)

Answer 9

Solution: If we multiply each number by itself, the value will change, but not the relation of the numbers with respect to size. For example if a > b, then a^2 > b^2 and a^n > b^n. In other words, we can raise both sides of an inequality (or equation) to any positive power and the inequality (equality) will be preserved.

Therefore, if 10^1/10 > 3^1/3, then also (10^1/10)^30 > (3^1/3)^30.
Now the left side is (10^1/10)^30 = 10^3 = 1,000.
The right side is (3^1/3)^30 = 3^10. It is easy to see that this will be a lot greater than 1,000 - just start multiplying 3 × 3 × 3 × 3... by using paper and pencil and it won't take too long to reach a number that is more than 1,000. Or, simplify 3^10 as 9^5 and multiply 9 repeatedly by itself. If you calculate it completely, you will get 3^10 = 59,049.

Therefore, since 3^10 > 10^3, then also 3^1/3 > 10^1/10. You can check that using calculator: 3^1/3 ≈ 1.442 and 10^1/10 ≈ 1.258.

Answer 10

good job, you got it :)

Answer 11

Yes, I played with that idea, but decided it was too messy to explain

Answer 12

haha XD okay.

Answer 13

or maybe say this
x^10 = 10
y^3 = 3
y^9= 27

y times itself only 9 times gets us to a bigger number than x to itself *ten* times
so y is bigger than x
i.e. 3^(1/3) > 10^(1/10)