Please help! Maths problem

Parsley

This question is on the pre leaving for 2001 and, though it looks easy, it's very difficult. There's no power notation on this so i'll do one number to a line, ie if you write down the equation it'll be in the form x+y=129, with x and y being two to the power of x+2 and two to the power of 5-x respectively.

Please help, I'm desperate!!!

Solve for x:

2 to the power of (x+2)

plus

2 to the power of (5-x)

equals

129

pointywalnut

129 = 128 + 1.

1 = 2 to the power of 0.

You should be able to solve the rest of it yourself.

David19

Split 2^(x+2) into (2^x)(2^2) and similarly for 2^(5-x).

After you tidy it up let y = 2^x, you'll end up with a quadratic which you can solve for y from which you can then solve for 2^x.

Does that help?

Caryatnid

This is easy if you know the rules properly. Here's the answer, put PM me if you need more explanation.

Orig Sum: 2^(x+2) + 2^(5-x) = 129
Rewrite as: 2^x.2^2 + 2^5.2^(-x) = 129
Simplify: 2^x.4 + 32/2^x = 129
Multiply across by 2^x to get: 2^x.2^x.4 + 32 = 129.2^x
Rewrite as: 4(2^x)^x -129(2^x) + 32 = 0
Now, as the last post says, replace 2^x by y, to get: 4y^2 -129y + 32 = 0
Factorise, and get: (4y - 1)(y - 32) = 0
You can do it from here?
Substitute the 2^x back in to get x.


I get: x = -2 and x = 5.