I don't fully understand how "ln" works yet. I would really appreciate some help with this problem.
Enter all admissible solutions for:
ln(x+31)−ln(x+3)=ln(8)−ln(x).
$\displaystyle \ln(x+31) - \ln(x+3) = \ln(8) - \ln(x)$
using the difference property for logs ...
$\displaystyle \displaystyle \ln\left(\frac{x+31}{x+3}\right) = \ln\left(\frac{8}{x}\right)$
$\displaystyle \displaystyle \frac{x+31}{x+3} = \frac{8}{x}$
solve for x ... note that any solution for x must be greater than zero.