Question

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

Answer 1

`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)`

Answer 2

`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`