What is the true solution to the equation below?
$\ln e^{\ln x}+\ln e^{\ln x^2}=2 \ln 8$
To solve the equation \(\ln e^{\ln x} + \ln e^{\ln x^2} = 2 \ln 8\), we first define the domain. Since the natural logarithm \(\ln(u)\) is defined for \(u > 0\), we must have \(x > 0\) and \(x^2 > 0\). Thus, our solution must satisfy \(x > 0\). We recall two fundamental properties of logarithms and exponentials:
Using the property \(\ln e^A = A\), we simplify the terms on the left side of the equation:
\(\ln e^{\ln x} = \ln x\)
\(\ln e^{\ln x^2} = \ln x^2\)
Substituting these into the original equation, we get:
\(\ln x + \ln x^2 = 2 \ln 8\)
We can now apply the logarithmic product rule \(\ln A + \ln B = \ln(AB)\) to the left side:
\(\ln(x \cdot x^2) = \ln(x^3)\)
On the right side, we use the property \(n \ln A = \ln A^n\) to write:
\(2 \ln 8 = \ln(8^2) = \ln 64\)
Now, the equation is in the form:
\(\ln(x^3) = \ln 64\)
Since the natural logarithm function is one-to-one, we can equate the arguments of the logarithms:
\(x^3 = 64\)
Taking the cube root of both sides, we find:
\(x = \sqrt[3]{64} = 4\)
Since \(x=4 > 0\), it satisfies the original domain conditions. Comparing this to the given options, we select the second option, \(x = 4\).
Logarithmic equations are used in fields like chemistry to calculate the pH of solutions, where pH is defined as \(-\log_{10}[H^+]\). For example, if you know the concentration of hydrogen ions in a water sample and need to determine the acidity, you might encounter logarithmic equations similar to this one. By isolating variables through logarithmic properties, scientists can accurately measure and monitor environmental changes or chemical reactions.