Which of the following is a reasonable solution to the logarithmic equation ln(2−3x)=ln(−5x) ?(1 point)

Answer 1

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:

\[ \boxed{-1} \]

Answer 2

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)

Answer 3

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.

Answer 4

Use the Quotient Rule of Logarithms to simplify this expression showing your steps algebraically. log4 64−log4 16

Answer 5

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:

\[ \boxed{1} \]