How do solve \[\ln \left( x+6 \right)+\ln \left( x-6 \right)=0\]?

Hint: This type of problem is based on the concept of logarithm. First, we have to consider the left-hand side of the given equation. Using the rule of logarithm, that is \[\ln \left( a\cdot b \right)=\ln a+ \ln b\], we can write the LHS as \[\ln \left( \left( x+6 \right)\left( x-6 \right) \right)\]. Simplify further using the property \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]. Then consider the RHS. We know that \[\ln \left( 1 \right)=0\], substituting in the RHS we get an equation with logarithm on both sides of the equation. Take antilog to remove logarithm from both the sides of the equation. Add 36 to the RHS and LHS and simplify further. Take the square root on both sides of the equation to obtain the value of x which is the required answer.

Complete step-by-step solution:
According to the question, we are asked to solve the given equation \[\ln \left( x+6 \right)+\ln \left( x-6 \right)=0\].
We have been given the equation is \[\ln \left( x+6 \right)+\ln \left( x-6 \right)=0\]. -----(1)
Let us first consider the LHS.
LHS=\[\ln \left( x+6 \right)+\ln \left( x-6 \right)\]
We know that \[\ln \left( a\cdot b \right)=\ln a+ \ln b\]. Using this property of logarithm in the LHS, we get
LHS=\[\ln \left( \left( x+6 \right)\left( x-6 \right) \right)\]
Now, we need to consider the RHS.
RHS=0.
But we know that \[\ln \left( 1 \right)=0\].
Substituting this in the RHS, we get
RHS=\[\ln \left( 1 \right)\]
Therefore, the simplified equation (1) is
\[\ln \left( \left( x+6 \right)\left( x-6 \right) \right)=\ln \left( 1 \right)\] -----------(2)
In equation (2), we find that logarithm in both the sides of the equation.
On taking antilog on both the sides, we get the functions to which the logarithm is defined to be equal.
That is \[\left( x+6 \right)\left( x-6 \right)=1\].
Now, we need to use the formula \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\] to simplify the LHS of the obtained expression.
Here, a=x and b=6.
Therefore, \[\left( x+6 \right)\left( x-6 \right)={{x}^{2}}-{{6}^{2}}\].
We know that \[{{6}^{2}}=36\].
\[\Rightarrow \left( x+6 \right)\left( x-6 \right)={{x}^{2}}-36\]
Therefore, we get
\[{{x}^{2}}-36=1\]
Now, we have to add 36 on both the sides of the equation.
\[\Rightarrow {{x}^{2}}-36+36=1+36\]
\[\Rightarrow {{x}^{2}}-36+36=37\]
We know that terms with the same magnitude and opposite signs cancel out.
Therefore, we get
\[{{x}^{2}}=37\]
But we have to find the value of x.
So we have to take square root on both sides of the expression.
\[\Rightarrow \sqrt{{{x}^{2}}}=\sqrt{37}\]
But we know that \[\sqrt{{{x}^{2}}}=\pm x\]. Using this in the above expression, we get
\[x=\pm \sqrt{37}\]
We know that logarithm function is not defined for negative functions. Therefore, we should neglect the negative value of x.
\[\therefore x=\sqrt{37}\]
Hence, the values of x in the given equation \[\ln \left( x+6 \right)+\ln \left( x-6 \right)=0\] is \[\sqrt{37}\].

Note: Whenever you get this type of problem, we should always try to make the necessary changes in the given equation so that we can take antilog and cancel out the logarithm function from the equation. We should not consider \[-\sqrt{37}\] because logarithm function is not defined for negative values. We should avoid calculation mistakes based on sign conventions. Also don’t confuse with the rules of logarithm.