How do you solve #log(x^2+4)-log(x+2)=2+log(x-2)#?

1 Answer

Please see below.

Explanation:

Here,

#log(x^2+4)-log(x+2)=2+log(x-2)#

#=>log(x^2+4)-log(x+2)-log(x-2)=2#

#=>log(x^2+4)-{log(x+2)+log(x-2)}=2#

Using : #logM+logN=log(MN)#

#=>log(x^2+4)-log[(x+2)(x-2)]=2#

#=>log(x^2+4)-log(x^2-4)=2#

Using : #logM-logN=log(M/N)#

#log((x^2+4)/(x^2-4))=2#

#(i)#If it is common logarithm-logarithm to base #10# ,then

#log_10 ((x^2+4)/(x^2-4))=2#

#:.(x^2+4)/(x^2-4)=10^2,where,x^2!=4=>color(red)(x!=+-2#

#x^2+4=100(x^2-4)#

#:.100x^2-400-x^2-4=0#

#99x^2-404=0#

#:.x^2=404/99~~4.08#

#:.x=+-sqrt4.08~~2.02#

But, #x~~-2.02# will make #log(x-2)# meaningless.

#:. x~~2.02#

Note that most of the textbooks use #logx # as,
logarithm to base 10

#(ii)#If it is #color(blue)"natural logarithm ??-logarithm to base e "# ,then

#log_e ((x^2+4)/(x^2-4))=2#

#:.(x^2+4)/(x^2-4)=e^2,where,x^2!=4=>color(red)(x!=+-2#

#x^2+4=e^2(x^2-4)#

#:.e^2x^2-4e^2-x^2-4=0#

#:.e^2x^2-x^2=4e^2+4#

#:.x^2(e^2-1)=4(e^2+1)#

#:.x^2=4((e^2+1)/(e^2-1))#

#:.x=2sqrt((e^2+1)/(e^2-1)) #