Respuesta :
Hello,
1) Conditions:
x+2>=0==>x>-2
x>=0
So x>=0
1/5 ln(x+2)^5 +1/2 [ln x -ln(x^2+3x+2)^2]
=ln(x+2)+ ln√x - ln(x+1)(x+2)
[tex]=ln (x+2)+ln(\dfrac{ \sqrt{x} }{(x+1)(x+2)}) \\\\ = ln(\dfrac{ \sqrt{x} }{x+1} )[/tex]
1) Conditions:
x+2>=0==>x>-2
x>=0
So x>=0
1/5 ln(x+2)^5 +1/2 [ln x -ln(x^2+3x+2)^2]
=ln(x+2)+ ln√x - ln(x+1)(x+2)
[tex]=ln (x+2)+ln(\dfrac{ \sqrt{x} }{(x+1)(x+2)}) \\\\ = ln(\dfrac{ \sqrt{x} }{x+1} )[/tex]
Based on your question where to have the quantity to be express as a single logarithm, the given algorithm in your question must first to be analyze and based on my calculation and my simplyfication the answer is ln is equals to the quotients of sqrt of x over x plus 1.