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Sole the equation 2^x - x = 5
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at first you combine the like terms
Nope, the first step is to move the normal x over to the side. Giving you 2^x = x + 5, any idea what you do next? c:
I'm going to assume you meant \(\large x^2\) and not \(\large 2^x\) because that will get messy. To start off, you need to get everything to one side, so subtract 5: \(\large x^2-x\color{red}{-5} = 0 \) You can either try factoring, or using the quadratic formula: \(\huge x = \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-5)}}{2(1)} \) \(\Large x= \frac{1 \pm \sqrt{1+20}}{2} = \frac{1 \pm \sqrt{21}}{2} \) Which can be simplified to: \(\Large x = \frac{1 + \sqrt{21}}{2}, \frac{1 - \sqrt{21}}{2} \)
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