Thread: solving for x
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Old November 21st, 2008, 01:31 AM
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Quote:
Originally Posted by Kitty216 View Post
Solve for x



It looks so simple but when I tried it, I ended up getting strange numbers. Help!
Here's an outline - you can fill in the details.

You have (2^2)^{2x} - 2 \cdot (2^2)^{x+4} + (2^2)^8 = 0.

Substitute w = 2^x and do a bit of simplifying.

You end up with w^2 - 2^9 w + 2^{16} = 0.

This factorises: (w - 2^8)^2 = 0.

Therefore 2^x = 2^8.
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Last edited by mr fantastic; November 21st, 2008 at 01:45 AM.
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