Answer:
- domain: all terms are defined for all real numbers
- solution: x = 6
Step-by-step explanation:
Rewrite the equation as a single exponential. After taking the log, the solution becomes obvious.
[tex]5^{x-2}-7^{x-3}=7^{x-5}+11\cdot 5^{x-4}\\\\5^{x-2}-11\cdot 5^{x-4}=7^{x-3}+7^{x-5}\\\\5^{x-2}(1-11\cdot 5^{-2})=7^{x-3}(1+7^{-2})\\\\5^{x-2}\dfrac{14}{25}=7^{x-3}\dfrac{50}{49}\\\\1=\dfrac{2\cdot 7^{x-3}5^4}{2\cdot 5^{x-2}7^3}=\left(\dfrac{7}{5}\right)^{x-6}\\\\0=(x-6)\log(1.4)\\\\x=6[/tex]
