Why am I allowed to remove $\ln$ from both sides of an equation? When I have this equation: $\ln(x+2) = \ln(x^2)$, why can I just remove the $\ln$ from both sides by raising it to the power of e. Does this not permanently change what the equitation equals? Also, when I graph $x+2=x^2$ and $\ln(x+2) = \ln(x^2)$ the roots of the equation are the same. Why are they? 
Q1: Why can I remove the $\ln$ by raising both sides of the equation to e? Does this not permanently change what the equation is?
Q2: Why are the roots of the two equations above the same?
 A: Equations don't equal anything.  Equations are sentences.  You can replace a sentence with a different sentence as long as it is also true.  1)$27x = 54y$ is a sentence.  2)$x = 2y$ is a different sentence.  The first sentence in only true if the second one is too and vice versa.  But they are different sentences.  We can replace one with the other because one follows from the other.  We don't care what $27x$ equals.  We care what $x$ equals.
If $a = b$ then $anythinginvolving(a) = anythinginvolving(b)$.  So if $a = b$ then $e^a = e^b$.  Likewise $a + 27 = b+ 27$ or $a^2 - \sqrt{3a} + 5 =  b^2 -\sqrt{3b} + 5$.  If $a = b$ we can do anything to either side.
So if $\ln (x+2) = \ln x^2$ we can do $e^{\ln(x+2)} = e^{\ln x^2}$.  Why not?  If we wanted to, we could say $\ln(x+2)$ fried in batter and stuffed inside $\sqrt{\text{a tomato}}=\ln(x^2)$ fried in batter and stuffed inside $\sqrt{\text{a tomato}}$ for all anyone cares.
But $e^{\ln(x+2)}= e^{\ln x^2}$ is a smart thing to care about.  (Whereas  $\ln(x+2)$ fried in batter and stuffed inside $\sqrt{\text{a tomato}}=\ln(x^2)$ fried in batter and stuffed inside $\sqrt{\text{a tomato}}$  is an utterly stupid thing to care about.)
BY DEFINITION:  $e^{\ln a} = a$, ALWAYS.   And for real numbers $\ln e^a = a$.  $e^x$ and $\ln x$ are inverses and that means they can "undo" each other.  Ith this way they are just like multiplication and division.  If you have $3(x+2) = 3x^2$ you can "undo" the "times $3$ by dividing both sides by $3$ because division "undoes" multiplication.  
So $\ln(x+2) = \ln x^2$ is a true sentence.  We don't care.
That means $e^{\ln(x+2)} = e^{\ln x^2}$ is a different true sentence.
But this different true sentence means the same thing as:
$x+2 = e^{\ln(x+1)} = e^{\ln x^2} = x^2$ so now we have yet another true sentence:
$x + 2= x^2$.  And we care because the takes us one step closer to soling what $x$ is.
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Postscipt:
If $a = b$ we can do anything to either side and get a true statement.
BUt be careful.  Not everything we do can be undone.
For example:  If $x = 7$ then $3x + 5 = 3*7 + 5 = 26$.  And $0\times (3x+5) = 0\times 26$ and so $0 = 0$.  These are all true sentences.  But the are not equivalent sentences.  Multiplying by $0$ can not be "undone".  But multiplying by $3$ can be undone (by dividing by $3$) and adding $5$ can be undone (by subtracting $5$).
Raising $e$ to a power can be undone by taking the logorithm.  And taking the logorithm can be undone by raising $e$ to the power but only if the number was positive to begin with.
A: Q1: We have $$\ln g(x) = \ln f(x) \implies \ln g(x) - \ln f(x) = 0 \implies \ln \dfrac{g(x)}{f(x)}=0 \implies \dfrac{g(x)}{f(x)}=1.$$
Q2: Note that the previous operations are only valid if $\ln g(x)$ and $\ln f(x)$ are defined. We get into trouble if the arguments are allowed to be negative or zero. The roots are the same because the solution to the equation happens to be in the allowed region. 
A: Q1. Because of injectivity of the log.
Q2. Same answer as Q1
A: The mapping
\begin{array}{cccc}
   \ln: &(0,\infty) &\to & \mathbb R \\
      & x &\mapsto &\ln(x)
\end{array}
is one-to-one onto, and strictly increasing.
Similarly, the mapping 
\begin{array}{cccc}
   \exp: & \mathbb R &\to &(0,\infty) \\
      & x &\mapsto &\exp(x)=e^x
\end{array}
is one-to-one onto, and strictly increasing.


*
  
*$x < y \iff e^x < e^y$
  
*$x = y \iff e^x = e^y$
  
*$x < y \iff \ln(x) < \ln(y)$
  
*$x = y \iff \ln(x) = \ln(y)$
