Index Misc Questions

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  • Solve 2^{x+1}+\sqrt{2^{x}}=1--Senghong79 04:56, 2 July 2010 (UTC)
    • Answer x=-2\,
    • Solution (Guided)
      • If you have no idea where to start, do look back at the lessons, and try it again on your own first. Keep in mind that if you are able to get an answer, a simple check will ensure whether it is correct or wrong
      • Analyze :
        • Base : Equal bases Can we rearrange to 2^{m}=2^{n}\,?
        • Number of terms : More than two So we CAN'T. Even if we take the trouble to change it to {\color{Blue}2}^{x+1}+{\color{Blue}2}^{\frac{1}{2}x}={\color{Blue}2}^{0}, there is still no way to change it to 2^{m}=2^{n}\,
        • Method to use : Substitution, {\color{Blue}2}^{{\color{Blue}x}+1}+\sqrt{{\color{Blue}2^{x}}}=1, so we use u=2^{x}\,
      • 2^{x+1}+\sqrt{2^{x}}=1 Let's prepare it for substitution
      • 2\left(2^{x}\right)+\sqrt{2^{x}}=1
      • \mbox{Let }u=2^{x}\,
      • 2u+\sqrt{u}=1
        • How would we deal with the \sqrt{u}?
        • We would need to square both sides, but before that, we must rearrange so that the square root is on one side and everything else is on the other side
      • \sqrt{u}=1-2u
      • u= \left(1-2u\right)^{2}
      • u= 1-4u+4u^{2}\, just a quadratic
      • 4u^{2}-5u+1 = 0 \,
      • \left(4u-1\right)\left(u-1\right)=0
      • u=\frac{1}{4} \mbox{ or } u=1 Finished? No. Find x\,
      • 2^{x}=\frac{1}{4} \mbox{ or } 2^{x}=1 \,
      • x=-2 \mbox{ or } x=0 \, Finished? NO.
      • Let try check quickly (mental calculation or calculator) : x=-2, 2^{-1}+\sqrt{\frac{1}{4}}=\frac{1}{2}+\frac{1}{2}=1OK
        • x=0, 2^{1}+\sqrt{2^{0}}=2+1=3 NOT OK!
        • Did we do anything wrong? Careless mistakes?
        • No. So why is it wrong? What steps did we take?
        • We squared both sides, which might introduce an extra answer which is wrong (refer to lesson to know why)
        • So we would actually need to write the checking as part of the solution and reject any solutions which do not satisfy the question
      • \mbox{When } x=-2, \mbox{L.H.S } = \frac{1}{2}+\sqrt{\frac{1}{4}}=0 = \mbox{R.H.S }
      • \mbox{When } x=1, \mbox{L.H.S } = 2+\sqrt{1}=1 \neq \mbox{R.H.S }
      • \therefore x = -2
    • Solution
      • \begin{align} & 2^{x+1}+\sqrt{2^{x}}=1 \\ & 2\left(2^{x}\right)+\sqrt{2^{x}}=1 \\ & \mbox{Let }u=2^{x}\\ & 2u+\sqrt{u}=1\\ & \sqrt{u}=1-2u \\ & u= \left(1-2u\right)^{2} \\ & u= 1-4u+4u^{2} \\ & 4u^{2}-5u+1 = 0 \\ & \left(4u-1\right)\left(u-1\right)=0\\ & u=\frac{1}{4} \mbox{ or } u=1 \\ & 2^{x}=\frac{1}{4} \mbox{ or } 2^{x}=1 \\ & x = -2 \mbox{ or } x=0 \\ & \mbox{When } x=-2, \mbox{L.H.S } = \frac{1}{2}+\sqrt{\frac{1}{4}}=0 = \mbox{R.H.S } \\ & \mbox{When } x=1, \mbox{L.H.S } = 2+\sqrt{1}=1 \neq \mbox{R.H.S } \\ & \therefore x = -2 \\ \end{align}
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