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

\(\displaystyle{2}^{{2}}{x}−{10}{\left({2}^{{x}}\right)}+{16}={0}\)

\(\displaystyle{\left({2}^{{x}}\right)}^{{2}}−{10}{\left({2}^{{x}}\right)}+{16}={0}\)

\(\displaystyle{y}^{{2}}−{10}{y}+{16}={0}\)

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