Notes: `2^(2x)-12*2^x+32=0` Solve the equation.

Sunday, 7 May 2017

$\displaystyle{{2}}^{{{2}{x}}}-{12}\cdot{{2}}^{{x}}+{32}={0}$ Solve the equation.

We are asked to solve $\displaystyle{{2}}^{{{2}{x}}}-{12}\cdot{{2}}^{{x}}+{32}={0}$ :

Rewrite as $\displaystyle{{\left({{2}}^{{x}}\right)}}^{{2}}-{12}\cdot{{2}}^{{x}}+{32}={0}$ and let $\displaystyle{y}={{2}}^{{x}}$ ; then

$\displaystyle{{y}}^{{2}}-{12}{y}+{32}={0}$ and

(y-8)(y-4)=0 so y=8 or y=4.

y=8 ==> $\displaystyle{{2}}^{{x}}={8}=\Rightarrow{x}={3}$

y=4 ==> $\displaystyle{{2}}^{{x}}={4}=\Rightarrow{x}={2}$

Checking we find that these are indeed solutions.

The graph:

We are asked to solve $\displaystyle{{2}}^{{{2}{x}}}-{12}\cdot{{2}}^{{x}}+{32}={0}$ :

Rewrite as $\displaystyle{{\left({{2}}^{{x}}\right)}}^{{2}}-{12}\cdot{{2}}^{{x}}+{32}={0}$ and let $\displaystyle{y}={{2}}^{{x}}$ ; then

$\displaystyle{{y}}^{{2}}-{12}{y}+{32}={0}$ and

(y-8)(y-4)=0 so y=8 or y=4.

y=8 ==> $\displaystyle{{2}}^{{x}}={8}=\Rightarrow{x}={3}$

y=4 ==> $\displaystyle{{2}}^{{x}}={4}=\Rightarrow{x}={2}$

Checking we find that these are indeed solutions.

The graph:

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