Inequalities question (1 Viewer)

Aviator_13 · Jan 22, 2018

2^2x − 2(2^x) ≤ −1

What methods can I use to solve this??

InteGrand · Jan 22, 2018

Aviator_13 said:
2^2x − 2(2^x) ≤ −1

What methods can I use to solve this??

$\noindent \textbf{Hint.} If you substitute $u = 2^{x}$, then you get a quadratic inequation to solve in $u$ (noting that $u > 0$ also because $u = 2^{x}$). So you can do the question if you've learnt how to solve quadratic inequations.$

captainhelium · Jan 22, 2018

This is what I got.

$2^{2x} -2(2^x) \leq -1$

$Let u= 2^x$

$∴ Using substitution...$

$u^2 - 2u \leq -1$

$u^2 -2u + 1 \leq 0$

$(u-1)^2 \leq 0$

$When you graph the above inequality, you'll see that the parabola has a value equal to 0 at u = 1. It's never less than 0 though. Therefore, all we need to do is just substitute u = 1 back into u = 2^x$

$2^x = 1$

$x = \log_{2}1$

I hope that's correct!

Aviator_13 · Jan 22, 2018

Thank you!!

Inequalities question (1 Viewer)

Aviator_13

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InteGrand

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captainhelium

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Aviator_13

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