# (2^2x)−10(2^x)+16=0 (Hint: Let y=2^x).

Question
Equations
$$\displaystyle{\left({2}^{{2}}{x}\right)}−{10}{\left({2}^{{x}}\right)}+{16}={0}$$ (Hint: Let $$\displaystyle{y}={2}^{{x}}$$).

2021-03-10
As hinted, let $$\displaystyle{y}={2}^{{x}}$$ so that the equation becomes:
$$\displaystyle{2}^{{2}}{x}−{10}{\left({2}^{{x}}\right)}+{16}={0}$$
$$\displaystyle{\left({2}^{{x}}\right)}^{{2}}−{10}{\left({2}^{{x}}\right)}+{16}={0}$$
$$\displaystyle{y}^{{2}}−{10}{y}+{16}={0}$$
Factor the left side:
(y−2)(y−8)=0
By zero product property,
y=2,8
Solve for xx by back-substituting $$\displaystyle{y}={2}^{{x}}$$
When y=2,
$$\displaystyle{2}^{{x}}={2}$$
$$\displaystyle{2}^{{x}}={2}^{{1}}$$
x=1
When y=8,
$$\displaystyle{2}^{{x}}={8}$$
$$\displaystyle{2}^{{x}}={2}^{{3}}$$
x=3

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