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

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

Question
Equations
asked 2021-03-09
\(\displaystyle{\left({2}^{{2}}{x}\right)}−{10}{\left({2}^{{x}}\right)}+{16}={0}\) (Hint: Let \(\displaystyle{y}={2}^{{x}}\)).

Answers (1)

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
0

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