Approximate all real values x such that 2ln2 - 2ln(x + 2) = ln(x - 5)

Question

Natural log has always confused me a bit, so problems like these tend to give me issues.

Answer 1

2ln2 - 2ln(x + 2) = ln(x - 5)

 

Note x > -2 and x > 5 to keep the arguments of the logarithms positive (x > 5 is sufficient).

 

2[ln2 - ln(x + 2)] = ln(x - 5)

 

2 ln[2/(x + 2)] = ln(x - 5)

 

ln{[2/(x + 2)]²} = ln(x - 5)

 

Take e⁽⁾ on both sides.

 

[2/(x + 2)]² = x - 5

 

4/(x + 2)² = x - 5

 

This will give a cubic equation in x to solve.  Keep in mind x > 5 is a restriction from the originally stated problem.

Answer 2

2ln2 - 2ln(x + 2) = ln(x - 5)

ln 4 - ln(x + 2) = ln(x + 5)

ln (4 / (x + 2)) = ln (x + 5)

4 / (x + 2) = x + 5

 

Can you solve for x?