Ln(x^2+4)=2lnx+ln4 Can someone help?

2 Answers
Jan 27, 2018

#x=2/sqrt(3)#

Explanation:

#Ln(x^2+4)=2lnx+ln4#

by property of logarithms

#2lnx=lnx^2#

#lnx^2+ln4=ln4x^2#

then

#Ln(x^2+4)=2lnx+ln4 -> ln(x^2+4)=ln4x^2#

removing the logarithm

#x^2+4=4x^2#

#4x^2=x^2+4#

#4x^2-x^2=4#

#3x^2=4#

#x^2=4/3#

#x=2/sqrt(3)#

Jan 27, 2018

#x=+-(2sqrt3)/3#

A lot of method explanation given so you can see where everything comes from.

Explanation:

#color(blue)("The teaching bit")#
If every thing on both sides of the equals is logs then with a b it of luck you can 'undo' the logs and deal with just the numbers and variables.

#color(brown)("What you do to one side of the equals you also do to the other")#
Example : #ln(x^2)=ln(16)#

Using a very old term you take the antilogarithm of both sides
Now they tend to use 'inverse log' written as #ln^-1("some value")#

So for the given example we have:

#ln^(-1)(x^2)=ln^(-1)(16) color(white)("d") ->color(white)("d")x^2=16#

#color(white)("dddddddddddddddddd")->color(white)("d")x=+-4#

It is #+-4# as #(-4)xx(-4)=+16=(+4)xx(+4)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#
Note: adding logs is consequential to the multiplication of the source values.

#2xx6 ->ln(2)+ln(6)# then you take the inverse log of the answer/

Note that #ln(x^2)# may be written as #2ln(x)#. It has the same value

Given:
#color(green)( color(white)("ddddddd") ln(x^2+4)=color(red)(2)ln(x)+ln(4))#

This is the same as

# color(green)(color(white)("ddddddd")ln(x^2+4)=ln(x^color(red)(2))+ln(4))#

#color(green)( color(white)("ddddddd")ln(x^2+4)=ln(x^2xx4))#

Take loginverse of both sides

#color(green)( color(white)("ddddddd")x^2+4=x^2xx4)#

# color(green)(color(white)("ddddddd")x^2+4=4x^2)#

Subtract #color(red)(x^2)# from both sides

#color(green)( color(white)("ddddddd")x^2color(red)(-x^2)+4=4x^2color(red)(-x^2)#

#color(green)(color(white)("ddddddddd.ddddd")4=3x^2)#

Divide both sides by #color(red)(3)#

#color(green)(color(white)("ddddddddd.ddddd")4/color(red)(3)=3/color(red)(3)x^2)#

#color(green)(color(white)("dddddddddddddd")4/3=x^2#

#color(green)(x=+-sqrt(4/3)color(white)("dd")->color(white)("dd")x=+-sqrt4/sqrt3color(white)("dd")->color(white)("dd")x=+-2/sqrt3)#

It is considered bad practice to have a root in the denominator so you should remove it if possible. Multiply by 1 and you do not change the value. However, 1 comes in many forms.

#color(green)(x=+-[2/sqrt3color(red)(xx1)]color(white)("ddd")->color(white)("ddd")x=+-[2/sqrt3color(red)(xxsqrt3/sqrt3)])#

#color(green)(color(white)("ddddddddddddddddd")->color(white)("ddd")x=+-(2sqrt3)/3)#

#color(red)("There is a problem with this. See the comment by Georgs")#

So we have #x=+(2sqrt(3))/3#