Answer:
No real solutions.
Step-by-step explanation:
We want to solve the equation:
[tex]\ln(x-2)+\ln (x+2)=4[/tex]
Recall that:
[tex]\displaystyle \ln a-\ln b=\ln\left(\frac{a}{b}\right)[/tex]
Therefore:
[tex]\displaystyle \ln\left(\frac{x-2}{x+2}\right)=4[/tex]
By Definition:
[tex]\displaystyle e^4=\frac{x-2}{x+2}[/tex]
Cross-multiply:
[tex]e^4(x+2)=x-2[/tex]
Distribute:
[tex]xe^4+2e^4=x-2[/tex]
Isolate the x:
[tex]xe^4-x=-2-2e^4[/tex]
Factor:
[tex]\displaystyle x(e^4-1)=-2-2e^4[/tex]
Divide. Therefore:
[tex]\displaystyle x=\frac{-2-2e^4}{e^4-1}=\frac{2+2e^4}{1-e^4}\approx-2.07[/tex]
However, note that logs cannot be negative and must be nonzero. According to the first logarithm, x > 2 and according to the second, x > -2. Since the answer is not greater than two, there are no real solutions.
