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}#
