# How do you solve (1) ^x = 3^(x-6)?

May 17, 2016

Real solution:

$x = 6$

Complex solutions:

$x = 6 + \frac{2 k \pi i}{\ln} 3$ for any integer $k \in \mathbb{Z}$

#### Explanation:

For any value of $x$, we have ${1}^{x} = 1$

So our equation simplifies to:

${3}^{x - 6} = 1$

If $x = 6$ then:

${3}^{x - 6} = {3}^{0} = 1$

satisfying the equation.

As a Real valued function of Reals $f \left(t\right) = {3}^{t}$ is one to one, so this is the unique Real solution.

Complex solutions

If $k$ is any integer then:

$1 = {e}^{2 k \pi i} = {\left({e}^{\ln 3}\right)}^{\frac{2 k \pi i}{\ln 3}} = {3}^{\frac{2 k \pi i}{\ln 3}}$

So:

${3}^{t + \frac{2 k \pi i}{\ln} 3} = {3}^{t} \cdot {3}^{\frac{2 k \pi i}{\ln} 3} = {3}^{t}$

Hence the equation ${\left(1\right)}^{x} = {3}^{x - 6}$ has solutions:

$x = 6 + \frac{2 k \pi i}{\ln} 3$

for any integer $k \in \mathbb{Z}$