Dive into the world of knowledge and get your queries resolved at IDNLearn.com. Join our Q&A platform to receive prompt and accurate responses from knowledgeable professionals in various fields.
Sagot :
To determine the transformed logarithmic equation that passes through the given points [tex]\((1,0)\)[/tex] and [tex]\((0,3)\)[/tex] using the equation [tex]\( y = a \log ( \pm x + c) \)[/tex], we will follow these steps.
### Step 1: Set Up the General Equation
The general form of the equation is:
[tex]\[ y = a \log (\pm x + c) \][/tex]
### Step 2: Substitute the First Point [tex]\((1,0)\)[/tex] into the Equation
For the point [tex]\((1,0)\)[/tex], substitute [tex]\( x = 1 \)[/tex] and [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = a \log (\pm 1 + c) \][/tex]
This simplifies to:
[tex]\[ 0 = a \log (\pm 1 + c) \][/tex]
Since [tex]\( a \neq 0 \)[/tex], this implies:
[tex]\[ \log (\pm 1 + c) = 0 \][/tex]
[tex]\[ \pm 1 + c = 1 \][/tex]
Therefore, [tex]\( c = 1 - 1 = 0 \)[/tex] or [tex]\( c = 1 + 1 = 2 \)[/tex].
So we get two possibilities:
[tex]\[ \pm 1 + c = 1 \Rightarrow c = 0 \quad \text{or} \quad c = 2 \][/tex]
### Step 3: Substitute the Second Point [tex]\((0,3)\)[/tex] into the Equation
For the point [tex]\((0,3)\)[/tex], substitute [tex]\( x = 0 \)[/tex] and [tex]\( y = 3 \)[/tex]:
[tex]\[ 3 = a \log (\pm 0 + c) \][/tex]
This simplifies to:
[tex]\[ 3 = a \log (c) \][/tex]
### Step 4: Solve for Constants [tex]\( a \)[/tex] and [tex]\( c \)[/tex]
#### Case 1: [tex]\( c = 0 \)[/tex]
If [tex]\( c = 0 \)[/tex], the logarithmic term [tex]\( \log(c) \)[/tex] is undefined since [tex]\(\log(0)\)[/tex] is not defined. So, [tex]\( c = 0 \)[/tex] is not a valid solution.
#### Case 2: [tex]\( c = 2 \)[/tex]
If [tex]\( c = 2 \)[/tex], then we have:
[tex]\[ 3 = a \log (2) \][/tex]
[tex]\[ a = \frac{3}{\log 2} \][/tex]
### Step 5: Write the Final Equation
So, substituting [tex]\( a \)[/tex] and [tex]\( c \)[/tex] back into the general form, we get:
[tex]\[ y = \frac{3}{\log 2} \log (x + 2) \][/tex]
Therefore, the equation of the transformed logarithm that passes through the points [tex]\((1, 0)\)[/tex] and [tex]\((0, 3)\)[/tex] is:
[tex]\[ f(x) = \frac{3}{\log 2} \log (x + 2) \][/tex]
### Step 1: Set Up the General Equation
The general form of the equation is:
[tex]\[ y = a \log (\pm x + c) \][/tex]
### Step 2: Substitute the First Point [tex]\((1,0)\)[/tex] into the Equation
For the point [tex]\((1,0)\)[/tex], substitute [tex]\( x = 1 \)[/tex] and [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = a \log (\pm 1 + c) \][/tex]
This simplifies to:
[tex]\[ 0 = a \log (\pm 1 + c) \][/tex]
Since [tex]\( a \neq 0 \)[/tex], this implies:
[tex]\[ \log (\pm 1 + c) = 0 \][/tex]
[tex]\[ \pm 1 + c = 1 \][/tex]
Therefore, [tex]\( c = 1 - 1 = 0 \)[/tex] or [tex]\( c = 1 + 1 = 2 \)[/tex].
So we get two possibilities:
[tex]\[ \pm 1 + c = 1 \Rightarrow c = 0 \quad \text{or} \quad c = 2 \][/tex]
### Step 3: Substitute the Second Point [tex]\((0,3)\)[/tex] into the Equation
For the point [tex]\((0,3)\)[/tex], substitute [tex]\( x = 0 \)[/tex] and [tex]\( y = 3 \)[/tex]:
[tex]\[ 3 = a \log (\pm 0 + c) \][/tex]
This simplifies to:
[tex]\[ 3 = a \log (c) \][/tex]
### Step 4: Solve for Constants [tex]\( a \)[/tex] and [tex]\( c \)[/tex]
#### Case 1: [tex]\( c = 0 \)[/tex]
If [tex]\( c = 0 \)[/tex], the logarithmic term [tex]\( \log(c) \)[/tex] is undefined since [tex]\(\log(0)\)[/tex] is not defined. So, [tex]\( c = 0 \)[/tex] is not a valid solution.
#### Case 2: [tex]\( c = 2 \)[/tex]
If [tex]\( c = 2 \)[/tex], then we have:
[tex]\[ 3 = a \log (2) \][/tex]
[tex]\[ a = \frac{3}{\log 2} \][/tex]
### Step 5: Write the Final Equation
So, substituting [tex]\( a \)[/tex] and [tex]\( c \)[/tex] back into the general form, we get:
[tex]\[ y = \frac{3}{\log 2} \log (x + 2) \][/tex]
Therefore, the equation of the transformed logarithm that passes through the points [tex]\((1, 0)\)[/tex] and [tex]\((0, 3)\)[/tex] is:
[tex]\[ f(x) = \frac{3}{\log 2} \log (x + 2) \][/tex]
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.