Question

what is the true solution to $3\ln 2 + \ln 8 = 2\ln(4x)$?
$\bigcirc$ $x = 1$
$\bigcirc$ $x = 2$
$\bigcirc$ $x = 4$
$\bigcirc$ $x = 8$

Explanation:

Step1: Simplify left - hand side using logarithm properties

Recall the logarithm property \(n\ln a=\ln(a^{n})\) and \(\ln a+\ln b = \ln(ab)\).
First, for \(3\ln2\), by the property \(n\ln a=\ln(a^{n})\), we have \(3\ln2=\ln(2^{3})=\ln8\).
Then the left - hand side of the equation \(3\ln2+\ln8\) becomes \(\ln8+\ln8\).
Using the property \(\ln a+\ln b=\ln(ab)\), \(\ln8+\ln8=\ln(8\times8)=\ln64\).

Step2: Simplify right - hand side using logarithm properties

For the right - hand side \(2\ln(4x)\), by the property \(n\ln a = \ln(a^{n})\), we get \(2\ln(4x)=\ln((4x)^{2})=\ln(16x^{2})\).

Step3: Solve the resulting logarithmic equation

Now our equation is \(\ln64=\ln(16x^{2})\).
Since the natural logarithm function \(y = \ln x\) is one - to - one (if \(\ln a=\ln b\), then \(a = b\) for \(a>0,b>0\)), we can set the arguments equal to each other:
\(64 = 16x^{2}\)
Divide both sides of the equation by 16: \(\frac{64}{16}=x^{2}\), so \(4 = x^{2}\)
Take the square root of both sides. But we also need to consider the domain of the original logarithmic function. The argument of \(\ln(4x)\) must be positive, so \(4x>0\) or \(x > 0\).
From \(x^{2}=4\), we have \(x=\pm2\), but since \(x>0\), we consider \(x = 2\) (Wait, let's check our simplification again. Wait, we made a mistake in step 1. Let's re - do step 1.

Wait, original left - hand side: \(3\ln2+\ln8\). \(3\ln2=\ln(2^{3})=\ln8\), so \(3\ln2+\ln8=\ln8+\ln8=\ln(8\times8)=\ln64\) is correct. Right - hand side: \(2\ln(4x)=\ln((4x)^{2})=\ln(16x^{2})\) is correct. Then \(\ln64=\ln(16x^{2})\) implies \(64 = 16x^{2}\), \(x^{2}=\frac{64}{16}=4\), \(x = 2\) (since \(x>0\) as \(4x>0\)). Let's verify by plugging \(x = 2\) into the original equation.

Left - hand side: \(3\ln2+\ln8=3\ln2+\ln2^{3}=3\ln2 + 3\ln2=6\ln2\)
Right - hand side: \(2\ln(4\times2)=2\ln8=2\times3\ln2 = 6\ln2\)
So \(x = 2\) is the solution.

Answer:

\(x = 2\) (The option is \(x = 2\) which is the second option: \(x = 2\))