Question

which expressions are equivalent to 2 ln a + 2 ln b - ln a? check all that apply. □ ln ab² - ln a □ ln a + 2 ln b □ ln a² + ln b² - ln a □ 2 ln ab □ ln ab² done

Explanation:

Step1: Simplify the original expression

The original expression is \(2\ln a + 2\ln b-\ln a\). Combine like terms: \((2\ln a-\ln a)+2\ln b=\ln a + 2\ln b\).

Step2: Analyze each option

• Option 1: \(\ln ab^{2}-\ln a\)

Using the property \(\ln x+\ln y = \ln(xy)\), \(\ln ab^{2}=\ln a+\ln b^{2}=\ln a + 2\ln b\) (since \(\ln x^{n}=n\ln x\)). Then \(\ln ab^{2}-\ln a=(\ln a + 2\ln b)-\ln a = 2\ln b\), which is not equivalent to the original simplified expression \(\ln a+2\ln b\)? Wait, no, wait. Wait the original expression after simplifying is \(\ln a + 2\ln b\). Wait, let's re - check. Wait the original expression: \(2\ln a+2\ln b-\ln a=(2\ln a-\ln a)+2\ln b=\ln a + 2\ln b\). Now for option 1: \(\ln(ab^{2})-\ln a\). First, \(\ln(ab^{2})=\ln a+\ln b^{2}=\ln a + 2\ln b\) (by \(\ln(xy)=\ln x+\ln y\) and \(\ln x^{n}=n\ln x\)). Then \(\ln(ab^{2})-\ln a=(\ln a + 2\ln b)-\ln a=2\ln b\)? No, that's wrong. Wait, no, I made a mistake. Wait the original expression: \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). Let's check option 2: \(\ln a+2\ln b\), which is exactly the simplified form of the original expression.

• Option 2: \(\ln a + 2\ln b\)

This is the simplified form of the original expression, so it is equivalent.

• Option 3: \(\ln a^{2}+\ln b^{2}-\ln a\)

Using \(\ln x^{n}=n\ln x\), \(\ln a^{2} = 2\ln a\) and \(\ln b^{2}=2\ln b\). So the expression becomes \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\), which is equivalent to the simplified original expression.

• Option 4: \(2\ln ab\)

Using \(\ln(xy)=\ln x+\ln y\) and \(\ln x^{n}=n\ln x\), \(2\ln ab = 2(\ln a+\ln b)=2\ln a + 2\ln b\), which is not equivalent to \(\ln a+2\ln b\).

• Option 5: \(\ln ab^{2}\)

Using \(\ln(xy)=\ln x+\ln y\) and \(\ln x^{n}=n\ln x\), \(\ln ab^{2}=\ln a+\ln b^{2}=\ln a + 2\ln b\), which is equivalent to the simplified original expression. Wait, wait, let's re - do the original expression simplification. Wait the original expression: \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). Now:

• Option 1: \(\ln(ab^{2})-\ln a\)

\(\ln(ab^{2})=\ln a+\ln b^{2}=\ln a + 2\ln b\), so \(\ln(ab^{2})-\ln a=(\ln a + 2\ln b)-\ln a = 2\ln b\), which is not equal to \(\ln a+2\ln b\). So option 1 is wrong.

• Option 2: \(\ln a + 2\ln b\): matches the simplified original expression.

• Option 3: \(\ln a^{2}+\ln b^{2}-\ln a\)

\(\ln a^{2}=2\ln a\), so \(2\ln a+\ln b^{2}-\ln a=\ln a+\ln b^{2}=\ln a + 2\ln b\) (since \(\ln b^{2}=2\ln b\)), which matches.

• Option 4: \(2\ln(ab)=2(\ln a+\ln b)=2\ln a + 2\ln b\), which is not equal to \(\ln a+2\ln b\).

• Option 5: \(\ln(ab^{2})=\ln a+\ln b^{2}=\ln a + 2\ln b\), which matches.

Wait, I think I made a mistake in the first analysis of option 1. Let's re - express the original expression:

Original expression: \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\)

Option 1: \(\ln(ab^{2})-\ln a\)

\(\ln(ab^{2})=\ln a+\ln(b^{2})=\ln a + 2\ln b\) (by \(\ln(xy)=\ln x+\ln y\) and \(\ln(x^{n})=n\ln x\))

Then \(\ln(ab^{2})-\ln a=(\ln a + 2\ln b)-\ln a=2\ln b\). Wait, that's not equal to \(\ln a + 2\ln b\). So option 1 is incorrect.

Option 2: \(\ln a+2\ln b\): correct.

Option 3: \(\ln a^{2}+\ln b^{2}-\ln a\)

\(\ln a^{2}=2\ln a\), so \(2\ln a+\ln b^{2}-\ln a=(2\ln a-\ln a)+\ln b^{2}=\ln a + 2\ln b\) (since \(\ln b^{2}=2\ln b\)): correct.

Option 4: \(2\ln(ab)=2(\ln a+\ln b)=2\ln a + 2\ln b\): not correct.

Option 5: \(\ln(ab^{2})=\ln a+\ln(b^{2})=\ln a + 2\ln b\): correct.

Answer:

B. \(\ln a + 2\ln b\), C. \(\ln a^{2}+\ln b^{2}-\ln a\), E. \(\ln ab^{2}\)