What are the potential solutions to the equation below?
$2 \ln (x+3)=0$
To solve the equation \(2 \ln (x+3) = 0\), we first identify the domain of the function. The natural logarithm \(\ln(y)\) is defined only for \(y > 0\). Therefore, for the term \(\ln(x+3)\) to be defined, we must have \(x+3 > 0\), which implies \(x > -3\).
Next, we isolate the logarithmic expression. Dividing both sides of the equation \(2 \ln (x+3) = 0\) by 2, we obtain:
\[\ln (x+3) = 0\]
Applying the inverse of the natural logarithm, which is the exponential function (base \(e\)), we have:
\[e^{\ln (x+3)} = e^0\]
\[x + 3 = 1\]
Solving the linear equation \(x + 3 = 1\) for \(x\), we subtract 3 from both sides:
\[x = 1 - 3\]
\[x = -2\]
We must now verify this result against our initial constraint \(x > -3\). Since \(-2 > -3\), the value \(x = -2\) is a valid solution. Note that the other value mentioned in some potential interpretations, \(x = -4\), does not satisfy the domain constraint \(x > -3\) because \(-4\) is not greater than \(-3\), and \(\ln(-4+3) = \ln(-1)\) is undefined in the real number system. Therefore, the only valid solution is \(x = -2\). Among the provided options, the only one that includes this correct solution is \(x = -2\) and \(x = -4\), which suggests that the original equation may have been intended as \(\ln((x+3)^2) = 0\). However, based strictly on the equation provided, only \(x = -2\) is valid.
Understanding logarithmic equations is crucial in fields like chemistry and biology, where pH levels (a measure of acidity or alkalinity) are calculated using logarithms: \(pH = -\log_{10}[H^+]\). If you know the pH level, you can use inverse logarithmic operations to find the concentration of hydrogen ions, \([H^+]\), in a solution. Solving equations like the one above helps in mastering the relationship between logarithmic forms and their algebraic counterparts, which is essential for working with exponential growth and decay models.