Answer :
Let's evaluate each expression to see if it is equivalent to the given expression [tex]\(2 \ln a + 2 \ln b - \ln a\)[/tex].
Given expression:
[tex]\[ 2 \ln a + 2 \ln b - \ln a \][/tex]
First, let's simplify the given expression step by step:
[tex]\[ 2 \ln a + 2 \ln b - \ln a \][/tex]
[tex]\[ = (2 \ln a - \ln a) + 2 \ln b \][/tex]
[tex]\[ = \ln a + 2 \ln b \][/tex]
Now, let's compare each provided expression with the simplified expression:
### 1. [tex]\(\ln(ab^2) - \ln a\)[/tex]
Simplify the expression:
[tex]\[ \ln(ab^2) - \ln a \][/tex]
Use the property of logarithms [tex]\( \ln(x) - \ln(y) = \ln\left(\frac{x}{y}\right) \)[/tex]:
[tex]\[ = \ln\left(\frac{ab^2}{a}\right) \][/tex]
[tex]\[ = \ln(b^2) \][/tex]
[tex]\[ = 2 \ln b \][/tex]
This is not equal to the simplified expression [tex]\( \ln a + 2 \ln b \)[/tex].
### 2. [tex]\(\ln a + 2 \ln b\)[/tex]
This matches exactly with our simplified expression [tex]\( \ln a + 2 \ln b \)[/tex]. So, this expression is equivalent.
### 3. [tex]\(\ln a^2 + \ln b^2 - \ln a\)[/tex]
Simplify the expression:
[tex]\[ \ln a^2 + \ln b^2 - \ln a \][/tex]
Use the property of logarithms [tex]\( \ln(x^2) = 2 \ln(x) \)[/tex]:
[tex]\[ = 2 \ln a + 2 \ln b - \ln a \][/tex]
This simplifies to:
[tex]\[ = (2 \ln a - \ln a) + 2 \ln b \][/tex]
[tex]\[ = \ln a + 2 \ln b \][/tex]
This is equal to the simplified expression [tex]\( \ln a + 2 \ln b \)[/tex]. So, this expression is equivalent.
### 4. [tex]\(2 \ln(ab)\)[/tex]
Simplify the expression:
[tex]\[ 2 \ln(ab) \][/tex]
Use the property of logarithms [tex]\( \ln(xy) = \ln x + \ln y \)[/tex]:
[tex]\[ = 2 (\ln a + \ln b) \][/tex]
[tex]\[ = 2 \ln a + 2 \ln b \][/tex]
This is equal to the given expression [tex]\(2 \ln a + 2 \ln b - \ln a\)[/tex]. However, when we bring back the subtraction of [tex]\( \ln a\)[/tex]:
[tex]\[ = 2 \ln a + 2 \ln b - \ln a \][/tex]
[tex]\[ = \ln a + 2 \ln b \][/tex]
So while the algebraic form matches after simplification, initially it may look different from the simplified final expression but they are indeed equivalent.
### 5. [tex]\( \ln(ab^2) \)[/tex]
Simplify the expression:
[tex]\[ \ln(ab^2) \][/tex]
Use the property of logarithms [tex]\( \ln(xy) = \ln x + \ln y \)[/tex]:
[tex]\[ = \ln a + \ln(b^2) \][/tex]
[tex]\[ = \ln a + 2 \ln b \][/tex]
This is equal to our simplified expression [tex]\( \ln a + 2 \ln b \)[/tex].
Summary of the given options:
- [tex]\(\ln(ab^2) - \ln a\)[/tex] is not equivalent.
- [tex]\(\ln a + 2 \ln b\)[/tex] is equivalent.
- [tex]\(\ln a^2 + \ln b^2 - \ln a\)[/tex] is equivalent.
- [tex]\(2 \ln(ab)\)[/tex] is not exactly equivalent because it directly equals [tex]\( 2 \ln a + 2 \ln b\)[/tex].
- [tex]\(\ln(ab^2)\)[/tex] is equivalent.
Thus, the expressions that are equivalent to [tex]\(2 \ln a + 2 \ln b - \ln a\)[/tex] are:
[tex]\[ \ln a + 2 \ln b \][/tex]
[tex]\[ \ln a^2 + \ln b^2 - \ln a \][/tex]
[tex]\[ \ln(ab^2) \][/tex]
Given expression:
[tex]\[ 2 \ln a + 2 \ln b - \ln a \][/tex]
First, let's simplify the given expression step by step:
[tex]\[ 2 \ln a + 2 \ln b - \ln a \][/tex]
[tex]\[ = (2 \ln a - \ln a) + 2 \ln b \][/tex]
[tex]\[ = \ln a + 2 \ln b \][/tex]
Now, let's compare each provided expression with the simplified expression:
### 1. [tex]\(\ln(ab^2) - \ln a\)[/tex]
Simplify the expression:
[tex]\[ \ln(ab^2) - \ln a \][/tex]
Use the property of logarithms [tex]\( \ln(x) - \ln(y) = \ln\left(\frac{x}{y}\right) \)[/tex]:
[tex]\[ = \ln\left(\frac{ab^2}{a}\right) \][/tex]
[tex]\[ = \ln(b^2) \][/tex]
[tex]\[ = 2 \ln b \][/tex]
This is not equal to the simplified expression [tex]\( \ln a + 2 \ln b \)[/tex].
### 2. [tex]\(\ln a + 2 \ln b\)[/tex]
This matches exactly with our simplified expression [tex]\( \ln a + 2 \ln b \)[/tex]. So, this expression is equivalent.
### 3. [tex]\(\ln a^2 + \ln b^2 - \ln a\)[/tex]
Simplify the expression:
[tex]\[ \ln a^2 + \ln b^2 - \ln a \][/tex]
Use the property of logarithms [tex]\( \ln(x^2) = 2 \ln(x) \)[/tex]:
[tex]\[ = 2 \ln a + 2 \ln b - \ln a \][/tex]
This simplifies to:
[tex]\[ = (2 \ln a - \ln a) + 2 \ln b \][/tex]
[tex]\[ = \ln a + 2 \ln b \][/tex]
This is equal to the simplified expression [tex]\( \ln a + 2 \ln b \)[/tex]. So, this expression is equivalent.
### 4. [tex]\(2 \ln(ab)\)[/tex]
Simplify the expression:
[tex]\[ 2 \ln(ab) \][/tex]
Use the property of logarithms [tex]\( \ln(xy) = \ln x + \ln y \)[/tex]:
[tex]\[ = 2 (\ln a + \ln b) \][/tex]
[tex]\[ = 2 \ln a + 2 \ln b \][/tex]
This is equal to the given expression [tex]\(2 \ln a + 2 \ln b - \ln a\)[/tex]. However, when we bring back the subtraction of [tex]\( \ln a\)[/tex]:
[tex]\[ = 2 \ln a + 2 \ln b - \ln a \][/tex]
[tex]\[ = \ln a + 2 \ln b \][/tex]
So while the algebraic form matches after simplification, initially it may look different from the simplified final expression but they are indeed equivalent.
### 5. [tex]\( \ln(ab^2) \)[/tex]
Simplify the expression:
[tex]\[ \ln(ab^2) \][/tex]
Use the property of logarithms [tex]\( \ln(xy) = \ln x + \ln y \)[/tex]:
[tex]\[ = \ln a + \ln(b^2) \][/tex]
[tex]\[ = \ln a + 2 \ln b \][/tex]
This is equal to our simplified expression [tex]\( \ln a + 2 \ln b \)[/tex].
Summary of the given options:
- [tex]\(\ln(ab^2) - \ln a\)[/tex] is not equivalent.
- [tex]\(\ln a + 2 \ln b\)[/tex] is equivalent.
- [tex]\(\ln a^2 + \ln b^2 - \ln a\)[/tex] is equivalent.
- [tex]\(2 \ln(ab)\)[/tex] is not exactly equivalent because it directly equals [tex]\( 2 \ln a + 2 \ln b\)[/tex].
- [tex]\(\ln(ab^2)\)[/tex] is equivalent.
Thus, the expressions that are equivalent to [tex]\(2 \ln a + 2 \ln b - \ln a\)[/tex] are:
[tex]\[ \ln a + 2 \ln b \][/tex]
[tex]\[ \ln a^2 + \ln b^2 - \ln a \][/tex]
[tex]\[ \ln(ab^2) \][/tex]
