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 = \frac{1}{a^b}
\frac{1}{10^{x-4}}=-2 , 10^{x-4}=-\frac{1}{2}
10^{x} * 10^{-4}=-\frac{1}{2}
10^{x} * \frac{1}{10^4}=-\frac{1}{2}
10^{x} = (10^4) * (-\frac{1}{2}) = -5000
10^{x} = \log_{10}(-5000) = \log_{10}((10^4) * (-\frac{1}{2}))
= 4 + \log_{10}(-1/2)
= 4 - [-1 \log_{10}(-1/2)]
= 4 - \log_{10}((-1/2)^-1)
= 4 - \log_{10}(-2)
Using change of base formula:
= 4-\frac{\ln(-2)}{\ln(10)}
using Euler's formula e^i\pi = -1 , i.e. ln(-1) = i\pi
= 4- \frac{\ln(2)+i\pi }{\ln(10)}
or expressed in rectangular form:
= 4- \frac{\ln(2)+i\pi }{\ln(10)}
4-\frac{\ln(2)}{\ln(10)}+\frac{\pi }{ln(10)}i
Thus,
x\approx 3.698970+1.364376354i
But since e^{2i\pi} = 1 and any number times 1 is itself,
\log_{10}(1) = \frac{2i\pi}{ln(10)}
and since 1^k = 1 for any integer k
\log_{10}(1) = \frac{2i\pi * k}{ln(10)} = \frac{2k\pi }{ln(10)}i
Thus all the solutions for x in the complex domain are:
x = 4-\frac{\ln(2)}{\ln(10)}+\frac{\pi +2k\pi }{ln(10)}i where k \in mathbb{Z} or
x\approx 3.698970+(1.364376354 + 2.728752708k)i where k \in mathbb{Z}
