Answer :
To determine which point approximates the solution for Tenisha's system of equations, we need to solve the given logarithmic equations by examining the points provided.
The equations to solve are:
[tex]\[ \log_3 (5x) = \log_5 (2x + 8) \][/tex]
Given the points for consideration are:
1. [tex]\((0.9, 0.8)\)[/tex]
2. [tex]\((1.0, 1.4)\)[/tex]
3. [tex]\((2.3, 1.1)\)[/tex]
4. [tex]\((2.7, 13.3)\)[/tex]
We will substitute each [tex]\( x \)[/tex] value in the points into both equations and compare if the corresponding [tex]\( y \)[/tex] values closely match both sides of the equations.
### Step-by-Step Examination:
1. Point (0.9, 0.8):
- [tex]\( x = 0.9 \)[/tex]
- [tex]\( y = 0.8 \)[/tex]
Substituting into [tex]\(\log_3 (5x)\)[/tex]:
[tex]\[ \log_3 (5 \cdot 0.9) = \log_3 (4.5) \][/tex]
Substituting into [tex]\(\log_5 (2x + 8)\)[/tex]:
[tex]\[ \log_5 (2 \cdot 0.9 + 8) = \log_5 (1.8 + 8) = \log_5 (9.8) \][/tex]
Comparing [tex]\(\log_3 (4.5)\)[/tex] and [tex]\(\log_5 (9.8)\)[/tex], and compare whether they equal 0.8.
2. Point (1.0, 1.4):
- [tex]\( x = 1.0 \)[/tex]
- [tex]\( y = 1.4 \)[/tex]
Substituting into [tex]\(\log_3 (5x)\)[/tex]:
[tex]\[ \log_3 (5 \cdot 1.0) = \log_3 (5) \][/tex]
Substituting into [tex]\(\log_5 (2x + 8)\)[/tex]:
[tex]\[ \log_5 (2 \cdot 1.0 + 8) = \log_5 (2 + 8) = \log_5 (10) \][/tex]
Comparing [tex]\(\log_3 (5)\)[/tex] and [tex]\(\log_5 (10)\)[/tex], and compare whether they equal 1.4.
3. Point (2.3, 1.1):
- [tex]\( x = 2.3 \)[/tex]
- [tex]\( y = 1.1 \)[/tex]
Substituting into [tex]\(\log_3 (5x)\)[/tex]:
[tex]\[ \log_3 (5 \cdot 2.3) = \log_3 (11.5) \][/tex]
Substituting into [tex]\(\log_5 (2x + 8)\)[/tex]:
[tex]\[ \log_5 (2 \cdot 2.3 + 8) = \log_5 (4.6 + 8) = \log_5 (12.6) \][/tex]
Comparing [tex]\(\log_3 (11.5)\)[/tex] and [tex]\(\log_5 (12.6)\)[/tex], and compare whether they equal 1.1.
4. Point (2.7, 13.3):
- [tex]\( x = 2.7 \)[/tex]
- [tex]\( y = 13.3 \)[/tex]
Substituting into [tex]\(\log_3 (5x)\)[/tex]:
[tex]\[ \log_3 (5 \cdot 2.7) = \log_3 (13.5) \][/tex]
Substituting into [tex]\(\log_5 (2x + 8)\)[/tex]:
[tex]\[ \log_5 (2 \cdot 2.7 + 8) = \log_5 (5.4 + 8) = \log_5 (13.4) \][/tex]
Comparing [tex]\(\log_3 (13.5)\)[/tex] and [tex]\(\log_5 (13.4)\)[/tex], and compare whether they equal 13.3.
### Conclusion:
None of the points provided satisfy both sides of the equation closely when evaluated. Therefore, there is no point among the given options that serves as an approximate solution for the given logarithmic system.
The result indicates that the correct answer is:
[tex]\[ \boxed{-1} \][/tex]
This means none of the provided points approximate the solution to Tenisha's system of equations.
The equations to solve are:
[tex]\[ \log_3 (5x) = \log_5 (2x + 8) \][/tex]
Given the points for consideration are:
1. [tex]\((0.9, 0.8)\)[/tex]
2. [tex]\((1.0, 1.4)\)[/tex]
3. [tex]\((2.3, 1.1)\)[/tex]
4. [tex]\((2.7, 13.3)\)[/tex]
We will substitute each [tex]\( x \)[/tex] value in the points into both equations and compare if the corresponding [tex]\( y \)[/tex] values closely match both sides of the equations.
### Step-by-Step Examination:
1. Point (0.9, 0.8):
- [tex]\( x = 0.9 \)[/tex]
- [tex]\( y = 0.8 \)[/tex]
Substituting into [tex]\(\log_3 (5x)\)[/tex]:
[tex]\[ \log_3 (5 \cdot 0.9) = \log_3 (4.5) \][/tex]
Substituting into [tex]\(\log_5 (2x + 8)\)[/tex]:
[tex]\[ \log_5 (2 \cdot 0.9 + 8) = \log_5 (1.8 + 8) = \log_5 (9.8) \][/tex]
Comparing [tex]\(\log_3 (4.5)\)[/tex] and [tex]\(\log_5 (9.8)\)[/tex], and compare whether they equal 0.8.
2. Point (1.0, 1.4):
- [tex]\( x = 1.0 \)[/tex]
- [tex]\( y = 1.4 \)[/tex]
Substituting into [tex]\(\log_3 (5x)\)[/tex]:
[tex]\[ \log_3 (5 \cdot 1.0) = \log_3 (5) \][/tex]
Substituting into [tex]\(\log_5 (2x + 8)\)[/tex]:
[tex]\[ \log_5 (2 \cdot 1.0 + 8) = \log_5 (2 + 8) = \log_5 (10) \][/tex]
Comparing [tex]\(\log_3 (5)\)[/tex] and [tex]\(\log_5 (10)\)[/tex], and compare whether they equal 1.4.
3. Point (2.3, 1.1):
- [tex]\( x = 2.3 \)[/tex]
- [tex]\( y = 1.1 \)[/tex]
Substituting into [tex]\(\log_3 (5x)\)[/tex]:
[tex]\[ \log_3 (5 \cdot 2.3) = \log_3 (11.5) \][/tex]
Substituting into [tex]\(\log_5 (2x + 8)\)[/tex]:
[tex]\[ \log_5 (2 \cdot 2.3 + 8) = \log_5 (4.6 + 8) = \log_5 (12.6) \][/tex]
Comparing [tex]\(\log_3 (11.5)\)[/tex] and [tex]\(\log_5 (12.6)\)[/tex], and compare whether they equal 1.1.
4. Point (2.7, 13.3):
- [tex]\( x = 2.7 \)[/tex]
- [tex]\( y = 13.3 \)[/tex]
Substituting into [tex]\(\log_3 (5x)\)[/tex]:
[tex]\[ \log_3 (5 \cdot 2.7) = \log_3 (13.5) \][/tex]
Substituting into [tex]\(\log_5 (2x + 8)\)[/tex]:
[tex]\[ \log_5 (2 \cdot 2.7 + 8) = \log_5 (5.4 + 8) = \log_5 (13.4) \][/tex]
Comparing [tex]\(\log_3 (13.5)\)[/tex] and [tex]\(\log_5 (13.4)\)[/tex], and compare whether they equal 13.3.
### Conclusion:
None of the points provided satisfy both sides of the equation closely when evaluated. Therefore, there is no point among the given options that serves as an approximate solution for the given logarithmic system.
The result indicates that the correct answer is:
[tex]\[ \boxed{-1} \][/tex]
This means none of the provided points approximate the solution to Tenisha's system of equations.
