# Question #93891

Jan 24, 2018

$x = 2$

#### Explanation:

Your notation is very ambiguous. I am assuming the equation is
${3}^{2 x} - 2 \cdot {3}^{x + 2} + 81 = 0$. If not, then this has all been in vain.
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${3}^{2 x} - 2 \cdot {3}^{x + 2} + 81 = 0$

We can rewrite this using the laws of indices:

${3}^{2 x} = {\left({3}^{x}\right)}^{2}$

$- 2 \cdot {3}^{x + 2} = - 2 \cdot {3}^{x} \cdot {3}^{2}$

So we have:

${\left({3}^{x}\right)}^{2} - 2 \cdot {3}^{x} \cdot {3}^{2} + 81 = 0$

Simplify:

${\left({3}^{x}\right)}^{2} - 18 \cdot {3}^{x} + 81 = 0$

This is a quadratic in ${3}^{x}$

Let $u = {3}^{x}$

Then:

${u}^{2} - 18 u + 81 = 0$

Factor:

${\left(u - 9\right)}^{2} = 0 \implies u = 9$

But:

$u = {3}^{x} \implies {3}^{x} = 9$

${3}^{x} = {3}^{2} \implies x = 2$