# Hello friends, please, I need to know how to get the answer out of this exponential equation? thank you

## Nov 5, 2016

$x = 3$

#### Explanation:

First, we will use the properties of exponents that

• ${a}^{x + y} = {a}^{x} {a}^{y}$
• ${\left({a}^{x}\right)}^{y} = {a}^{x y} = {\left({a}^{y}\right)}^{x}$

${9}^{x - 1} - 2 \cdot {3}^{x} - 27 = 0$

$\implies {9}^{- 1} \cdot {9}^{x} - 2 \cdot {3}^{x} - 27 = 0$

$\implies \frac{1}{9} \cdot {\left({3}^{2}\right)}^{x} - 2 \cdot {3}^{x} - 27 = 0$

$\implies \frac{1}{9} \cdot {\left({3}^{x}\right)}^{2} - 2 \cdot {3}^{x} - 27 = 0$

With these manipulations, we can see this has the form of a quadratic equation, but with ${3}^{x}$ as the variable rather than just $x$. As such, we will apply the quadratic formula:

${3}^{x} = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \left(\frac{1}{9}\right) \left(- 27\right)}}{2 \left(\frac{1}{9}\right)}$

$= \frac{2 \pm \sqrt{16}}{\frac{2}{9}}$

$= \frac{2 \pm 4}{\frac{2}{9}}$

$= 9 \pm 18$

Note that ${3}^{x} > 0$ for all real $x$. Thus the case in which ${3}^{x} = 9 - 18 = - 9$ has no real solutions. This leaves us with the case in which ${3}^{x} = 9 + 18$.

${3}^{x} = 9 + 18$

$\implies {3}^{x} = 27$

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

The answer is pretty clear at this point, but we can justify it properly by using logarithms. We will use the following properties of logarithms:

• $\log \left({a}^{x}\right) = x \log \left(a\right)$
• ${\log}_{a} \left(a\right) = 1$

With those, we have

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

$\implies x {\log}_{3} \left(3\right) = 3 {\log}_{3} \left(3\right)$

$\therefore x = 3$

Finally, we can check our answer:

${9}^{3 - 1} - 2 \cdot {3}^{3} - 27 = 81 - 2 \cdot 27 - 27 = 0$

as desired.