Question
Which of the following is a reasonable solution to the logarithmic equation ln(2−3x)=ln(−5x) ?(1 point)
Answers
GPT-4o mini
To solve the equation \( \ln(2 - 3x) = \ln(-5x) \), we can use properties of logarithms. Since both sides are equal, we can exponentiate both sides to eliminate the logarithm:
\[
2 - 3x = -5x
\]
Now, let's solve for \( x \):
\[
2 - 3x = -5x \\
2 = -5x + 3x \\
2 = -2x \\
x = -1
\]
Now we need to check if this solution is valid by substituting \( x = -1 \) back into the original logarithmic expressions to ensure they remain defined and positive:
1. Substitute \( x = -1 \) into \( 2 - 3x \):
\[
2 - 3(-1) = 2 + 3 = 5 \quad (\text{This is positive.})
\]
2. Substitute \( x = -1 \) into \( -5x \):
\[
-5(-1) = 5 \quad (\text{This is also positive.})
\]
Since all logarithmic arguments are positive at \( x = -1 \), this solution is reasonable.
Thus, the solution to the equation \( \ln(2 - 3x) = \ln(-5x) \) is:
\[
\boxed{-1}
\]
\[
2 - 3x = -5x
\]
Now, let's solve for \( x \):
\[
2 - 3x = -5x \\
2 = -5x + 3x \\
2 = -2x \\
x = -1
\]
Now we need to check if this solution is valid by substituting \( x = -1 \) back into the original logarithmic expressions to ensure they remain defined and positive:
1. Substitute \( x = -1 \) into \( 2 - 3x \):
\[
2 - 3(-1) = 2 + 3 = 5 \quad (\text{This is positive.})
\]
2. Substitute \( x = -1 \) into \( -5x \):
\[
-5(-1) = 5 \quad (\text{This is also positive.})
\]
Since all logarithmic arguments are positive at \( x = -1 \), this solution is reasonable.
Thus, the solution to the equation \( \ln(2 - 3x) = \ln(-5x) \) is:
\[
\boxed{-1}
\]
Joy used the change of base formula to rewrite log520 as log20log5 . Payne rewrote it as ln20ln5 . Who is correct? Payne, Joy, both, neither? Explain how you know.(4 points)
GPT-4o mini
To determine who is correct, let's look at how the change of base formula works for logarithms.
The change of base formula states that:
\[
\log_b(a) = \frac{\log_k(a)}{\log_k(b)}
\]
where \( b \) is the base, \( a \) is the argument, and \( k \) can be any positive number (commonly \( 10 \) or \( e \)).
### Joy's Rewrite:
Joy rewrote \( \log_5(20) \) as \( \frac{\log(20)}{\log(5)} \).
Assuming she used base 10 (with log as base 10) or any base (common logs), her expression is correct if interpreted as:
\[
\log_5(20) = \frac{\log(20)}{\log(5)}
\]
This can be correctly interpreted as using any base for logs (including base 10).
### Payne's Rewrite:
Payne rewrote \( \log_5(20) \) as \( \frac{\ln(20)}{\ln(5)} \).
This is also accurate as it applies the change of base formula using \( e \) as the base:
\[
\log_5(20) = \frac{\ln(20)}{\ln(5)}
\]
### Conclusion
Both rewrites by Joy and Payne are valid applications of the change of base formula, since both expressions correctly represent the change from the base \( 5 \) to some other base (where Joy uses a general logarithm and Payne uses natural logarithm).
Thus, the answer is:
**Both Joy and Payne are correct.**
The change of base formula states that:
\[
\log_b(a) = \frac{\log_k(a)}{\log_k(b)}
\]
where \( b \) is the base, \( a \) is the argument, and \( k \) can be any positive number (commonly \( 10 \) or \( e \)).
### Joy's Rewrite:
Joy rewrote \( \log_5(20) \) as \( \frac{\log(20)}{\log(5)} \).
Assuming she used base 10 (with log as base 10) or any base (common logs), her expression is correct if interpreted as:
\[
\log_5(20) = \frac{\log(20)}{\log(5)}
\]
This can be correctly interpreted as using any base for logs (including base 10).
### Payne's Rewrite:
Payne rewrote \( \log_5(20) \) as \( \frac{\ln(20)}{\ln(5)} \).
This is also accurate as it applies the change of base formula using \( e \) as the base:
\[
\log_5(20) = \frac{\ln(20)}{\ln(5)}
\]
### Conclusion
Both rewrites by Joy and Payne are valid applications of the change of base formula, since both expressions correctly represent the change from the base \( 5 \) to some other base (where Joy uses a general logarithm and Payne uses natural logarithm).
Thus, the answer is:
**Both Joy and Payne are correct.**
Use the Quotient Rule of Logarithms to simplify this expression showing your steps algebraically. log4 64−log4 16
GPT-4o mini
To simplify the expression \( \log_4 64 - \log_4 16 \) using the Quotient Rule of Logarithms, we can follow these steps:
### Step 1: Apply the Quotient Rule
The Quotient Rule states that for any logarithm \(\log_b(a) - \log_b(c)\), you can combine the two logarithms into one:
\[
\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)
\]
Here, we will apply the Quotient Rule to the expression:
\[
\log_4 64 - \log_4 16 = \log_4 \left(\frac{64}{16}\right)
\]
### Step 2: Simplify the Fraction
Next, simplify the fraction \( \frac{64}{16} \):
\[
\frac{64}{16} = 4
\]
So our expression now reads:
\[
\log_4 (4)
\]
### Step 3: Evaluate the Logarithm
The logarithm \( \log_4 (4) \) asks "To what power must 4 be raised to obtain 4?" The answer is:
\[
\log_4 (4) = 1
\]
### Final Result
Thus, the simplified expression is:
\[
\log_4 64 - \log_4 16 = 1
\]
In conclusion, the final, simplified result is:
\[
\boxed{1}
\]
### Step 1: Apply the Quotient Rule
The Quotient Rule states that for any logarithm \(\log_b(a) - \log_b(c)\), you can combine the two logarithms into one:
\[
\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)
\]
Here, we will apply the Quotient Rule to the expression:
\[
\log_4 64 - \log_4 16 = \log_4 \left(\frac{64}{16}\right)
\]
### Step 2: Simplify the Fraction
Next, simplify the fraction \( \frac{64}{16} \):
\[
\frac{64}{16} = 4
\]
So our expression now reads:
\[
\log_4 (4)
\]
### Step 3: Evaluate the Logarithm
The logarithm \( \log_4 (4) \) asks "To what power must 4 be raised to obtain 4?" The answer is:
\[
\log_4 (4) = 1
\]
### Final Result
Thus, the simplified expression is:
\[
\log_4 64 - \log_4 16 = 1
\]
In conclusion, the final, simplified result is:
\[
\boxed{1}
\]