Q. Identify the solution set of \(3 \ln(4) = 2 \ln(x)\).

Q: How do I solve 3\ln 4 = 2\ln x?

Use power rules: 3\ln(4)=\ln(4^3)=\ln(64) and 2\ln(x)=\ln(x^2). So \ln(64)=\ln(x^2) \Rightarrow x^2=64. With domain x>0, solution is x=8.

Q: Why can't x=-8 be accepted?

The natural log \ln(x) is defined only for x > 0. Although x^2 = 64 gives \pm 8, -8 is extraneous because \ln(-8) is undefined in real numbers.

Q: Could I solve it by exponentiating both sides?

Yes. From 3\ln 4=2\ln x rewrite as \ln 64=\ln(x^2) or exponentiate: e^{\ln 64}=e^{\ln(x^2)} giving 64=x^2, then x=8 (positive).

Q: Which logarithm properties are used?

Mainly the power rule: a\ln b=\ln(b^a). Also equality rule: if \ln A=\ln B then A=B (for positive A,B).

Q: What if the logs were base 10 instead of natural log?

Same method applies: 3\log 4=2\log x leads to \log(4^3)=\log(x^2), so 64=x^2 and x=8 (positive). Base doesn't matter if both logs use the same base.

Q: Are there common mistakes to avoid?

Yes: forgetting domain x>0, dropping the power rule and incorrectly dividing logs, or accepting negative root from x^2 without checking \ln domain.

Q: What's the graphical interpretation?

Graph y=3\ln 4 (horizontal line) and y=2\ln x. Their intersection gives the solution x=8, the unique positive x where 2\ln x equals the constant 3\ln 4.

Answer

Divide both sides by 2: \( \ln x = \tfrac{3}{2}\ln 4 \).
Exponentiate: \( x = e^{\tfrac{3}{2}\ln 4} = 4^{3/2} = (\sqrt{4})^{3} = 2^{3} = 8. \)
Solution set: \(\{8\}\).

Detailed Explanation

Problem

Identify the solution set of the equation

\(3\ln 4 = 2\ln x\).

Step-by-step solution with detailed explanations

Recognize the domain: The natural logarithm function \(\ln x\) is defined only for \(x>0\). Therefore any solution must satisfy

\(x>0\).
Use a logarithm power rule to simplify one side: The rule \(k\ln a = \ln(a^k)\) (valid for \(a>0\)) allows us to rewrite the left-hand side:

\(3\ln 4 = \ln(4^3) = \ln 64\).
Express the equation with equal logarithms: Substituting the simplification gives

\(\ln 64 = 2\ln x\).

Use the power rule on the right-hand side as well: \(2\ln x = \ln(x^2)\) provided \(x>0\). Thus

\(\ln 64 = \ln(x^2)\).
Remove the logarithms using injectivity of ln: Because the natural logarithm is one-to-one on its domain, if \(\ln A = \ln B\) and \(A>0\), \(B>0\), then \(A=B\). Applying this here gives

\(64 = x^2\).
Solve the resulting algebraic equation: Solve \(x^2 = 64\). This yields

\(x = 8\) or \(x = -8\).
Apply the domain restriction: Recall from step 1 that we must have \(x>0\). Therefore \(x = -8\) is not admissible because \(\ln(-8)\) is undefined. The only valid solution is

\(x = 8\).
Optional check: Substitute \(x=8\) back into the original equation:

Left-hand side: \(3\ln 4 = 3(2\ln 2) = 6\ln 2\).

Right-hand side: \(2\ln 8 = 2(3\ln 2) = 6\ln 2\).

Both sides match, confirming the solution.

Solution set

\(\{8\}\)

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FAQs

How do I solve \(3\ln 4 = 2\ln x\)?

Use power rules: \(3\ln(4)=\ln(4^3)=\ln(64)\) and \(2\ln(x)=\ln(x^2)\). So \(\ln(64)=\ln(x^2) \Rightarrow x^2=64\). With domain \(x>0\), solution is \(x=8\).

Why can't \(x=-8\) be accepted?

The natural log \(\ln(x)\) is defined only for \(x > 0\). Although \(x^2 = 64\) gives \(\pm 8\), -8 is extraneous because \(\ln(-8)\) is undefined in real numbers.

Could I solve it by exponentiating both sides?

Yes. From \(3\ln 4=2\ln x\) rewrite as \(\ln 64=\ln(x^2)\) or exponentiate: \(e^{\ln 64}=e^{\ln(x^2)}\) giving \(64=x^2\), then \(x=8\) (positive).

Which logarithm properties are used?

Mainly the power rule: \(a\ln b=\ln(b^a)\). Also equality rule: if \(\ln A=\ln B\) then \(A=B\) (for positive \(A,B\)).

What if the logs were base 10 instead of natural log?

Same method applies: \(3\log 4=2\log x\) leads to \(\log(4^3)=\log(x^2)\), so \(64=x^2\) and \(x=8\) (positive). Base doesn't matter if both logs use the same base.

How can I check the solution quickly?

Are there common mistakes to avoid?

Yes: forgetting domain \(x>0\), dropping the power rule and incorrectly dividing logs, or accepting negative root from \(x^2\) without checking \(\ln\) domain.

What's the graphical interpretation?

Graph \(y=3\ln 4\) (horizontal line) and \(y=2\ln x\). Their intersection gives the solution \(x=8\), the unique positive x where \(2\ln x\) equals the constant \(3\ln 4\).

Find x in 3 ln 4 = 2 ln x precisely.
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