# Solve - Power to the Square

• Apr 17th 2007, 07:11 PM
thecapaccino
Solve - Power to the Square
Solve ( 2^1-sqrt 3) 1+sqrt 3.?

Again just in case you can see if properly. (2 squared to the power of 1 - the square root of 3) all to the square root of 1+ the square root of 3.

I think the answer is

2^ (1+sqrt3)+(1-sqrt3)
2^ 1
= 2

Is 2 the answer?
• Apr 17th 2007, 07:34 PM
ecMathGeek
Quote:

Originally Posted by thecapaccino
Solve ( 2^1-sqrt 3) 1+sqrt 3.?

Again just in case you can see if properly. (2 squared to the power of 1 - the square root of 3) all to the square root of 1+ the square root of 3.

I think the answer is

2^ (1+sqrt3)+(1-sqrt3)
2^ 1
= 2

Is 2 the answer?

I would like to confirm your answer, but the 2 different ways its written are unclear. I assume the problem you are trying to solve is:

2^[(1 + sqrt3) + (1 - sqrt3)] = 2^[2] = 4

OR

2^[(1 + sqrt3)*(1 - sqrt3)] = 2^[1 - 3] = 2^(-2) = 1/4

In either case, the answer is not 2
• Apr 17th 2007, 07:38 PM
ecMathGeek
Quote:

Originally Posted by thecapaccino
Solve ( 2^1-sqrt 3) 1+sqrt 3.?

Again just in case you can see if properly. (2 squared to the power of 1 - the square root of 3) all to the square root of 1+ the square root of 3.

I think the answer is

2^ (1+sqrt3)+(1-sqrt3)
2^ 1
= 2

Is 2 the answer?

To be clear, you've basically indicated about 4 (or 5) different problems:

1. (2^1 - sqrt3)*1 + sqrt3 = 2 - sqrt3 + sqrt3 = 2

2. (2^2^(1 - sqrt3))^[sqrt(1 + sqrt3)] ... which makes no sense

3. 2^(1 + sqrt3) + (1 - sqrt3) ... which also makes no sense

4. 2^[(1 + sqrt3)(1 - sqrt3)] = 2^(1 - sqrt3 + sqrt3 - 3) = 2^(-2) = 1/4
• Apr 17th 2007, 07:40 PM
thecapaccino
Just to confirm
Okay...I think the answer is 4.

My question is ....do you multiply or add if you have 1-sqrt3 ^ 1+sqrt 3?
• Apr 17th 2007, 07:43 PM
thecapaccino
To clear the original equation again
The first parenthesis has ( 2^1-sqrt 3). That then, is all to the sqrt of 1+sqrt 3. A bit better ?
• Apr 17th 2007, 07:54 PM
ecMathGeek
Quote:

Originally Posted by thecapaccino
The first parenthesis has ( 2^1-sqrt 3). That then, is all to the sqrt of 1+sqrt 3. A bit better ?

That's nothing like any of the things I did.

This would be easier to explain if I could explain it directly to you, but I'll do my best given that I'm doing this online:

2^1 - sqrt3 does not equal 2^(1 - sqrt3). According to Order of Operations, the first says first we do 2 raised to the first power, then we subract the square root of 3. In other words: 2^1 - sqrt3 = 2 - sqrt3 (and I doubt this is what your problem is. IF the problem is "2 raised to the power of 1 minus the square root of 3" then 2^(1 - sqrt3) is the correct way to write this.

sqrt of 1 + sqrt3 means sqrt(1 + sqrt3), which is a bit complicated to work with (and I doubt is the way the problem should be written). Using this, the problem would be:
[2^(1 - sqrt3)]^sqrt(1 + sqrt3) which is a very complicated problem to work with (and again, I doubt this is how your problem should look).

I hope this helps.

I'm going to venture a guess and assume your problem should actually look something like this:

[ 2^(1 - sqrt3) ]^(1 + sqrt3) ... does this seem correct?

If so, then the answer is 1/4
• Apr 17th 2007, 08:01 PM
thecapaccino
Closer
The last one seems correct. [ 2^(1 - sqrt3) ]^(1 + sqrt3) ... You are right.

It is the number 2 to the power of (1-sqrt3). And that equation all to the power of (1+sqrt3).

However, in order of operations...you would work on the exponents first...which means you would multiply (1-sqrt3) * (1+sqrt3). So the answer really is 1/4.

Thanks