Need help with an equation. I know the answer but I don't know the process of solving it.Can you help?

Question

so here goes the equation :

√3^p + 1 = 2^p.
I know the value of p is 2 .But what is the correct process of solving it?
thanks in advance.

Answer 1

I believe this problem must be solved by estimation, which is better done with a computer simulation that could be easily coded to find integer solutions, but here's how I would do it by hand...

Changing the equation around a bit:
sqrt(3^p) - 2^p = 1

Make a small guess and a large guess:

Small guess: p = -100
Large guess: p = 100
(in the computer program, the first small and large guesses would be the min and max binary numbers that can be represented on your computer)
p = -100 ---> LHS << 1
p = 100 ---> LHS << 0

Increase small guess by some Δx
Decrease large guess by Δx
Now since I know the answer, I am going to skip a few steps and make a very large increase and decrease in small and large p.

Small guess: p = 0
Large guess: p = 3
p=0 ---> LHS = 0
p=3 ---> LHS < 0

Small guess: p = 1
Large guess: p = 2
p=1 ---> LHS < 0
p=2 ---> LHS = 1

So when p=2, LHS = RHS = 1

Therefore p=2.

Answer 2

It would depend on what course you are taking.

If you are taking a calculus course, you are probably expected to use Newton's method to get an approximate answer, to whatever degree of accuracy you want.

If you are taking a course before calculus, you are probably expected to just try a few values for p and try to pin down how big the answer is. If p = 0 then the left side = 2 and the right side = 1, so the left side is bigger. But if p = 6, you will find that the right side is bigger. So somewhere between 0 and 6 there is a value for p where left side and right side are exactly equal.

It just so happens that by trial and error, you can easily find that exact value for p that works.

There is no algebraic method for solving this kind of equation in general (not like we have the quadratic formula for second degree polynomial equations).

Answer 3

Well....let's see....this is a hard one. You almost stumped me on this one. I tried solving for p, but with a square root in there it makes it hard so, if you square it, it cancels out the square root-- that means putting a 2 in for the p. You have to do that with the other p too. Since you used 2 use 2 on the other side.

3 + 1 = 2^2
4=4

Answer 4

You start by subbing 2 in for p.
(square root)3^2 + 1 = 2^2

Then you would do the exponents.

(square root) 9 + 1 = 4

Then you would find out the square root of 9 and add that to 1.

3 + 1 = 4
4 = 4

That is how to prove that it is true.

Answer 5

well, I could solve it if it was
√3^(p + 1) = 2^p, but then p would not equal 2. As it is, I'm stumped for now.

Answer 6

you are able to word that sqrt 3^p +a million <2^p while p>2 and sqrt 3^p+a million >2^p while 0

Answer 7

3^(p/2) - 2^p + 1 = 0
3^(p/2) = 2^p - 1 . . . . . .. squaring both
3^p = 2^(2p) - 2^(p+1) + 1
still can't get it . . . more time