Question

which expressions are equivalent to \\(\ln x + 2 \ln 5 + \ln 1\\)?
check all that apply.
\\(\square\\) \\(2 \ln 5x\\)
\\(\square\\) \\(\ln (x + 26)\\)
\\(\square\\) \\(\ln 25x + \ln 1\\)
\\(\square\\) \\(\ln 25x\\)
retry

Explanation:

Step1: Recall logarithm properties

We know that \( n\ln a=\ln a^n \) and \( \ln a+\ln b = \ln(ab) \), also \( \ln1 = 0 \).
First, simplify \( 2\ln5 \) using the power rule of logarithms: \( 2\ln5=\ln5^2=\ln25 \).

Step2: Simplify the original expression

The original expression is \( \ln x + 2\ln5+\ln1 \). Substitute \( 2\ln5=\ln25 \) and \( \ln1 = 0 \), we get:
\( \ln x+\ln25 + 0=\ln(x\times25)=\ln(25x) \) (using the product rule of logarithms \( \ln a+\ln b=\ln(ab) \))
Now let's check each option:

• Option 1: \( 2\ln(5x)=\ln(5x)^2=\ln(25x^2)

eq\ln(25x) \), so this is not equivalent.

• Option 2: \( \ln(x + 26) \) is a logarithm of a sum, while our simplified expression is a logarithm of a product, so they are not equivalent.
• Option 3: \( \ln(25x)+\ln1=\ln(25x)+0=\ln(25x) \), which is equivalent to our simplified expression (since \( \ln(25x)+\ln1=\ln(25x)+0=\ln(25x) \) and our original expression simplifies to \( \ln(25x) \)).
• Option 4: As we simplified the original expression to \( \ln(25x) \) (because \( \ln1 = 0 \)), this is equivalent.

Answer:

\( \ln 25x + \ln 1 \), \( \ln 25x \) (the third and fourth options, i.e., the option with \( \ln 25x+\ln 1 \) and the option with \( \ln 25x \))