can 5^sqrt(3) be solved without a calculator?

Yes.

how would you go about doing that?

5[sup]√3[/sup] is quite exact. If you go into decimals you will be getting inaccurate answers.

You can still get an approximate solution.

I suppose you could develop the Taylor series for 5^sqrt(x) and calculate its value at x = 3.

You could also think of a few other approximation tricks.
For example, we know that 5^sqrt(3) = sqrt3_root(125).
Maybe you have some approximation of sqrt3_root. (I doubt it's easier than what I said above though.)

The most obvious is that it's between 5 and 5^2 = 25.

It's also between 5^1.7 and 5^1.8, but I don't know how to calculate those.

You could combine two taylor series. Calculate x = sqrt(3) and then calculate 5^x @ that x. I don't think the second calculation would be easy though.

Just some ideas.

1.7 = 1 + 1/7
5[sup]1 + 1/7[/sup] = (5[sup]1[/sup])(5[sup]1/7[/sup])
5[sup]1/7[/sup] is about 1.26, so 5 * 1.26 = 6.3

1.7 = 1 + 1/8
5[sup]1 + 1/8[/sup] = (5[sup]1[/sup])(5[sup]1/8[/sup])
5[sup]1/8[/sup] is about 1.22, so 5 * 1.22 = 6.1

6.1 + 6.3 = 12.4
12.4 / 2 = 6.2
Thus your answer is about 6.2

[quote author=rabbit link=topic=14988.msg152629#msg152629 date=1147876001]
1.7 = 1 + 1/7
5[sup]1 + 1/7[/sup] = (5[sup]1[/sup])(5[sup]1/7[/sup])
5[sup]1/7[/sup] is about 1.26, so 5 * 1.26 = 6.3

1.7 = 1 + 1/8
5[sup]1 + 1/8[/sup] = (5[sup]1[/sup])(5[sup]1/8[/sup])
5[sup]1/8[/sup] is about 1.22, so 5 * 1.22 = 6.1

6.1 + 6.3 = 12.4
12.4 / 2 = 6.2
Thus your answer is about 6.2
[/quote]

5^sqrt(3) = 16.2424508

Well hey, nevermind :P

[quote author=Yoni link=topic=14988.msg152624#msg152624 date=1147868871]
I suppose you could develop the Taylor series for 5^sqrt(x) and calculate its value at x = 3.

You could also think of a few other approximation tricks.
For example, we know that 5^sqrt(3) = sqrt3_root(125).
Maybe you have some approximation of sqrt3_root. (I doubt it's easier than what I said above though.)

The most obvious is that it's between 5 and 5^2 = 25.

It's also between 5^1.7 and 5^1.8, but I don't know how to calculate those.

You could combine two taylor series. Calculate x = sqrt(3) and then calculate 5^x @ that x. I don't think the second calculation would be easy though.

Just some ideas.
[/quote]
thanks for the input. could you perhaps explain how you came up with 5^sqrt(3) = sqrt3_root(125)? thanks!

[quote author=rabbit link=topic=14988.msg152629#msg152629 date=1147876001]
1.7 = 1 + 1/7
1.7 = 1 + 1/8
[/quote]
umm ok.

[quote author=Probe link=topic=14988.msg152655#msg152655 date=1147915674]
thanks for the input. could you perhaps explain how you came up with 5^sqrt(3) = sqrt3_root(125)? thanks!
[/quote]
Yes.

sqrt(3) * sqrt(3) = 3
So,
sqrt(3) = 3 / sqrt(3)
And,
5^sqrt(3) = 5^(3 / sqrt(3)) = (5^3) ^ (1 / sqrt(3)) = 125^(1 / sqrt(3)) = sqrt3_root(125)

O shit!  It was early............