We are given:
\(\displaystyle{9}^{{x}}-{1}={27}^{{-{{x}}}}+{2}\)

Rewrite each base as a power of 3: \(\displaystyle{\left({3}^{{2}}\right)}^{{x}}-{1}={\left({3}^{{3}}\right)}^{{-{{x}}}}+{2}\)

Use the rule: \(\displaystyle{\left({a}^{{m}}\right)}^{{n}}={a}^{{m}}{n}\) \(\displaystyle{3}^{{{2}{\left({x}-{1}\right)}}}={3}^{{3}}^{\left(-{x}+{2}\right)}\)

Equate the exponents: 2(x-1)=3(-x+2)

Solve for x:

2x-2=-3x+8

2x=-3x+8

5x=8

\(\displaystyle{x}=\frac{{8}}{{5}}\) or 1.6

Rewrite each base as a power of 3: \(\displaystyle{\left({3}^{{2}}\right)}^{{x}}-{1}={\left({3}^{{3}}\right)}^{{-{{x}}}}+{2}\)

Use the rule: \(\displaystyle{\left({a}^{{m}}\right)}^{{n}}={a}^{{m}}{n}\) \(\displaystyle{3}^{{{2}{\left({x}-{1}\right)}}}={3}^{{3}}^{\left(-{x}+{2}\right)}\)

Equate the exponents: 2(x-1)=3(-x+2)

Solve for x:

2x-2=-3x+8

2x=-3x+8

5x=8

\(\displaystyle{x}=\frac{{8}}{{5}}\) or 1.6