# 9^x-1=27^-x+2

Question
$$\displaystyle{9}^{{x}}-{1}={27}^{{-{{x}}}}+{2}$$

2021-02-09
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

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