Question

which expressions are equivalent to $2\ln a + 2\ln b - \ln a$?
choose three correct answers.
$\ln ab^2 - \ln a$ $\ln ab^2$ $2\ln ab$
$\ln a + 2\ln b$ $\ln a^2 + \ln b^2 - \ln a$

Explanation:

Step1: Simplify the original expression

First, simplify \(2\ln a + 2\ln b-\ln a\). Combine like terms: \(2\ln a-\ln a + 2\ln b=\ln a + 2\ln b\). Also, using logarithm power rule \(n\ln x=\ln x^n\), \(2\ln a=\ln a^2\), \(2\ln b = \ln b^2\), so original expression is \(\ln a^2+\ln b^2-\ln a\).

Step2: Analyze each option

- Option 1: \(\ln ab^2-\ln a\). Using power rule, \(2\ln b=\ln b^2\), so \(2\ln a + 2\ln b=\ln a^2+\ln b^2=\ln a^2b^2\)? Wait, no, \(2\ln a+2\ln b=\ln a^2+\ln b^2=\ln(a^2b^2)\), but \(\ln ab^2=\ln a+\ln b^2\), and original expression after combining \(2\ln a-\ln a=\ln a\), so \(2\ln a + 2\ln b-\ln a=\ln a + 2\ln b=\ln a+\ln b^2=\ln(ab^2)\), and \(\ln ab^2-\ln a=\ln(ab^2)-\ln a=\ln(\frac{ab^2}{a})=\ln b^2\)? Wait, no, wait my mistake. Wait original expression: \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). Let's re - express \(\ln ab^2-\ln a\): \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), then \(\ln ab^2-\ln a=(\ln a + 2\ln b)-\ln a=2\ln b\)? No, that's wrong. Wait no, wait \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). Now, \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), so \(\ln ab^2-\ln a=(\ln a + 2\ln b)-\ln a=2\ln b\)? No, I messed up. Wait original expression: \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). Let's check \(\ln ab^2-\ln a\): \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), so \(\ln ab^2-\ln a=(\ln a + 2\ln b)-\ln a=2\ln b\)? No, that's not equal to original. Wait no, wait \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). Now, \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), so \(\ln ab^2\) is equal to \(\ln a + 2\ln b\) (since original is \(\ln a+2\ln b\)). Wait, no, original is \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), so \(\ln ab^2\) is equal to the simplified original expression. Wait, my earlier mistake: \(\ln ab^2-\ln a\): \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), then \(\ln ab^2-\ln a=(\ln a + 2\ln b)-\ln a=2\ln b\)? No, that's incorrect. Wait, no, \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). Let's take \(a = e\), \(b = e\). Original expression: \(2\ln e+2\ln e-\ln e=2 + 2-1 = 3\). \(\ln ab^2-\ln a=\ln(e\times e^2)-\ln e=\ln e^3-\ln e=3 - 1=2\)? No, that's not 3. Wait, I see my mistake. Wait \(2\ln a+2\ln b-\ln a=\ln a+2\ln b\). For \(a = e\), \(b = e\), \(\ln e+2\ln e=1 + 2=3\). \(\ln ab^2-\ln a=\ln(e\times e^2)-\ln e=\ln e^3-\ln e=3 - 1 = 2\), which is not 3. So my earlier analysis was wrong. Wait, let's start over.

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

Simplify \(E\): Combine like terms for \(\ln a\): \(2\ln a-\ln a=\ln a\), so \(E=\ln a + 2\ln b\)

Now, use power rule \(n\ln x=\ln x^n\):

- \(2\ln a=\ln a^2\), \(2\ln b=\ln b^2\), so \(E=\ln a^2+\ln b^2-\ln a\) (this is one equivalent form)

- Also, \(\ln a+2\ln b=\ln a+\ln b^2=\ln(ab^2)\) (using product rule \(\ln x+\ln y=\ln(xy)\))

- The expression \(\ln a + 2\ln b\) is the simplified form we got by combining like terms.

Now let's check each option:

1. \(\ln ab^2-\ln a\): \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), so \(\ln ab^2-\ln a=(\ln a + 2\ln b)-\ln a=2\ln b eq E\). Wait, no, wait if \(E=\ln a + 2\ln b\), and \(\ln ab^2=\ln a + 2\ln b\), so \(\ln ab^2\) is equal to \(E\), not \(\ln ab^2-\ln a\). Wait, I think I misread the option. The first option is \(\ln ab^2-\ln a\), the second is \(\ln ab^2\). Let's re - evaluate with \(a = e\), \(b = e\):

- Original \(E\): \(2\ln e+2\ln e-\ln e=2 + 2-1 = 3\)

- Option 1: \(\ln(e\times e^2)-\ln e=\ln e^3-\ln e=3 - 1 = 2 eq3\)

- Option 2: \(\ln(e\times e^2)=\ln e^3 = 3\), which equals \(E\)

- Option 3: \(2\ln(ab)\). If \(a = e\), \(b = e\), \(2\ln(e\times e)=2\ln e^2=2\times2 = 4 eq3\)

- Option 4: \(\ln a+2\ln b\). For \(a = e\), \(b = e\), \(\ln e+2\ln e=1 + 2 = 3\), equals \(E\)

- Option 5: \(\ln a^2+\ln b^2-\ln a\). For \(a = e\), \(b = e\), \(\ln e^2+\ln e^2-\ln e=2 + 2-1 = 3\), equals \(E\)

- Also, \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\) (which is \(E\)), so \(\ln ab^2\) is equivalent.

- \(\ln a + 2\ln b\) is the simplified \(E\)

- \(\ln a^2+\ln b^2-\ln a\) is equivalent as we saw.

Wait, earlier when I thought \(\ln ab^2-\ln a\) was wrong, but let's check the first option again. Wait, \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). \(\ln ab^2-\ln a=\ln a+\ln b^2-\ln a=\ln b^2\), which is not equal. But \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), which is equal to \(E\). So the correct options are:

- \(\ln ab^2\) (option 2)

- \(\ln a + 2\ln b\) (option 4)

- \(\ln a^2+\ln b^2-\ln a\) (option 5)

Wait, but also, let's check the first option again. Wait, \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\). \(\ln ab^2-\ln a=\ln a+\ln b^2-\ln a=\ln b^2\), no. Wait, maybe I made a mistake in the first option's analysis. Wait, \(2\ln a+2\ln b=\ln a^2+\ln b^2=\ln(a^2b^2)\), then \(2\ln a+2\ln b-\ln a=\ln(a^2b^2)-\ln a=\ln(\frac{a^2b^2}{a})=\ln(ab^2)\). Oh! Here's the mistake. I forgot that \(2\ln a=\ln a^2\), so \(2\ln a+2\ln b=\ln a^2+\ln b^2=\ln(a^2b^2)\), then subtract \(\ln a\): \(\ln(a^2b^2)-\ln a=\ln(\frac{a^2b^2}{a})=\ln(ab^2)\). Wait, no, \(2\ln a+2\ln b-\ln a=(2\ln a-\ln a)+2\ln b=\ln a + 2\ln b\), and \(\ln(ab^2)=\ln a+\ln b^2=\ln a + 2\ln b\), and \(\ln(a^2b^2)-\ln a=\ln a^2+\ln b^2-\ln a\). Also, \(\ln ab^2-\ln a\): Wait, \(\ln ab^2=\ln a+\ln b^2=\ln a + 2\ln b\), so \(\ln ab^2-\ln a=(\ln a + 2\ln b)-\ln a=2\ln b\), which is not equal. But \(\ln(ab^2)\) is equal to \(\ln a + 2\ln b\) (since \(\ln(ab^2)=\ln a+\ln b^2=\ln a + 2\ln b\)). And \(\ln a^2+\ln b^2-\ln a=2\ln a+\ln b^2-\ln a=\ln a+\ln b^2=\ln(ab^2)\), which is equal to \(E\). And \(\ln a + 2\ln b\) is equal to \(E\). Also, let's check the first option: \(\ln ab^2-\ln a\). Wait, \(2\ln a+2\ln b-\ln a=\ln a + 2\ln b\), and \(\ln ab^2-\ln a=\ln a+\ln b^2-\ln a=\ln b^2\), which is not equal. But wait, when we use the quotient rule on \(\ln(ab^2)-\ln a\), we get \(\ln(\frac{ab^2}{a})=\ln b^2\), which is not equal to \(E\). But when we simplify \(2\ln a+2\ln b-\ln a\) as \(\ln a^2+\ln b^2-\ln a\) (because \(2\ln a=\ln a^2\)), then \(\ln a^2+\ln b^2-\ln a\) is equivalent. Also, \(\ln a + 2\ln b\) is equivalent, and \(\ln ab^2\) is equivalent (since \(\ln a + 2\ln b=\ln a+\ln b^2=\ln(ab^2)\)).

So the three correct answers are:

- \(\ln ab^2\)

- \(\ln a + 2\ln b\)

- \(\ln a^2+\ln b^2-\ln a\)

Answer:

The three correct answers are:

• \(\boldsymbol{\ln ab^2}\)
• \(\boldsymbol{\ln a + 2\ln b}\)
• \(\boldsymbol{\ln a^2+\ln b^2-\ln a}\)