# How do you solve  10^(-x+4) +7 = 5?

May 15, 2018

$x \setminus \approx 3.698970 + \left(1.364376354 + 2.728752708 k\right) i$ where $k \setminus \in m a t h \boldsymbol{Z}$

#### Explanation:

No solutions in real domain, but there are in the complex domain:
First, subtract 7 from both sides to get:
${10}^{- x + 4} = - 2$
recall that ${a}^{-} b = \setminus \frac{1}{{a}^{b}}$
$\setminus \frac{1}{{10}^{x - 4}} = - 2$ , ${10}^{x - 4} = - \setminus \frac{1}{2}$
${10}^{x} \cdot {10}^{- 4} = - \setminus \frac{1}{2}$
${10}^{x} \cdot \setminus \frac{1}{{10}^{4}} = - \setminus \frac{1}{2}$
${10}^{x} = \left({10}^{4}\right) \cdot \left(- \setminus \frac{1}{2}\right) = - 5000$
${10}^{x} = \setminus {\log}_{10} \left(- 5000\right) = \setminus {\log}_{10} \left(\left({10}^{4}\right) \cdot \left(- \setminus \frac{1}{2}\right)\right)$

$= 4 + \setminus {\log}_{10} \left(- \frac{1}{2}\right)$
$= 4 - \left[- 1 \setminus {\log}_{10} \left(- \frac{1}{2}\right)\right]$
$= 4 - \setminus {\log}_{10} \left({\left(- \frac{1}{2}\right)}^{-} 1\right)$
$= 4 - \setminus {\log}_{10} \left(- 2\right)$

Using change of base formula:
$= 4 - \setminus \frac{\setminus \ln \left(- 2\right)}{\setminus \ln \left(10\right)}$
using Euler's formula ${e}^{i} \setminus \pi = - 1$ , i.e. $\ln \left(- 1\right) = i \setminus \pi$
$= 4 - \setminus \frac{\setminus \ln \left(2\right) + i \setminus \pi}{\setminus \ln \left(10\right)}$
or expressed in rectangular form:
$= 4 - \setminus \frac{\setminus \ln \left(2\right) + i \setminus \pi}{\setminus \ln \left(10\right)}$
$4 - \setminus \frac{\setminus \ln \left(2\right)}{\setminus \ln \left(10\right)} + \setminus \frac{\setminus \pi}{\ln \left(10\right)} i$
Thus,
$x \setminus \approx 3.698970 + 1.364376354 i$

But since ${e}^{2 i \setminus \pi} = 1$ and any number times 1 is itself,
$\setminus {\log}_{10} \left(1\right) = \setminus \frac{2 i \setminus \pi}{\ln \left(10\right)}$
and since ${1}^{k} = 1$ for any integer k
$\setminus {\log}_{10} \left(1\right) = \setminus \frac{2 i \setminus \pi \cdot k}{\ln \left(10\right)} = \setminus \frac{2 k \setminus \pi}{\ln \left(10\right)} i$

Thus all the solutions for x in the complex domain are:
$x = 4 - \setminus \frac{\setminus \ln \left(2\right)}{\setminus \ln \left(10\right)} + \setminus \frac{\setminus \pi + 2 k \setminus \pi}{\ln \left(10\right)} i$ where $k \setminus \in m a t h \boldsymbol{Z}$ or
$x \setminus \approx 3.698970 + \left(1.364376354 + 2.728752708 k\right) i$ where $k \setminus \in m a t h \boldsymbol{Z}$