Question

list the potential solutions to 2 ln x = 4 ln 2 from least to greatest.
x = and x =
done

Explanation:

Step1: Simplify the equation

Given \(2\ln x = 4\ln 2\), divide both sides by 2: \(\ln x=2\ln 2\)
Using the logarithm power rule \(a\ln b=\ln(b^a)\), we get \(\ln x = \ln(2^2)=\ln 4\)

Step2: Solve for x

Since \(\ln\) is a one - to - one function, if \(\ln a=\ln b\), then \(a = b\) (for \(a>0,b > 0\)). So from \(\ln x=\ln 4\), we have \(x = 4\) or we can also consider the domain of the natural logarithm. The equation \(2\ln x=4\ln 2\) can also be rewritten by exponentiating both sides with base \(e\).
Starting from \(2\ln x=4\ln 2\), we can also use the property of logarithms in another way. First, rewrite the equation as \(\ln x^{2}=\ln 2^{4}\) (using \(a\ln b=\ln(b^{a})\)). Then \(x^{2}=16\) (since \(\ln\) is one - to - one). Solving \(x^{2}=16\) gives \(x = 4\) or \(x=- 4\). But we must remember that the domain of \(y = \ln x\) is \(x>0\), so \(x=-4\) is not in the domain of the original function \(y = \ln x\). Wait, there is a mistake in the previous thought. Let's go back.
The original equation is \(2\ln x=4\ln 2\). The domain of \(\ln x\) requires \(x>0\). Let's solve it correctly:
Divide both sides of \(2\ln x = 4\ln 2\) by 2: \(\ln x=2\ln 2=\ln(2^{2})=\ln 4\)
Since \(y = \ln x\) is injective (one - to - one) on \((0,+\infty)\), then \(x = 4\). But if we use the power rule on the left - hand side as \(\ln(x^{2})\) and the right - hand side as \(\ln(2^{4})\), we have \(x^{2}=16\), so \(x=\pm4\). But we need to check the domain of the original function. For \(\ln x\) to be defined, \(x>0\), so \(x = - 4\) is extraneous. Wait, but maybe the problem is considering the equation in a different way? No, the original equation is \(2\ln x\), so \(x>0\). But let's re - examine the equation:
\(2\ln x=4\ln 2\)
\(\ln x^{2}=\ln 16\) (because \(4\ln 2=\ln(2^{4})=\ln 16\))
Then \(x^{2}=16\), so \(x = 4\) or \(x=-4\). But since \(x>0\) (because of \(\ln x\)), \(x = 4\) is the solution. Wait, maybe the problem has a typo or I misread it. Wait, the equation is \(2\ln x=4\ln 2\). Let's solve it step by step:

1. Divide both sides by 2: \(\ln x = 2\ln 2\)
2. Use the power rule of logarithms: \(2\ln 2=\ln(2^{2})=\ln 4\), so \(\ln x=\ln 4\)
3. Since the natural logarithm function \(y = \ln x\) is one - to - one (if \(\ln a=\ln b\), then \(a = b\) for \(a,b>0\)), we have \(x = 4\). But if we consider the equation as \(2\ln|x|=4\ln 2\) (but the original equation is \(2\ln x\), not \(2\ln|x|\)), the domain of \(\ln x\) is \(x>0\), so \(x=-4\) is not a valid solution. Wait, maybe the problem was supposed to be \(2\ln|x| = 4\ln 2\). Let's assume that for a moment. Then:

• \(2\ln|x|=4\ln 2\)
• \(\ln|x| = 2\ln 2=\ln 4\)
• \(|x| = 4\)
• So \(x = 4\) or \(x=-4\)

Since the problem asks for potential solutions (maybe considering the algebraic solution before checking the domain), the potential solutions come from \(x^{2}=16\), so \(x=-4\) and \(x = 4\). And when we order them from least to greatest, we have \(x=-4\) and \(x = 4\).

Answer:

\(x=-4\) and \(x = 4\) (in the order from least to greatest, first \(x=-4\), then \(x = 4\))