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\\)

Explanation:

Step1: Simplify \(2\ln 5\)

Using the logarithm power rule \(a\ln b=\ln(b^a)\), we have \(2\ln 5 = \ln(5^2)=\ln 25\).

Step2: Simplify \(\ln 1\)

We know that \(\ln 1 = 0\) (since \(\log_a 1 = 0\) for any \(a>0,a
eq1\)).

Step3: Combine the logarithms

Using the logarithm product rule \(\ln a+\ln b=\ln(ab)\), the original expression \(\ln x + 2\ln 5+\ln 1\) becomes \(\ln x+\ln 25 + 0=\ln(25x)\) (because \(\ln x+\ln 25=\ln(25x)\)). Also, \(\ln 25x+\ln 1=\ln 25x + 0=\ln 25x\) (since \(\ln 1 = 0\)). Let's check the other options:

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

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

• For \(\ln(x + 26)\): This is a logarithm of a sum, not a product, so it's not equivalent to \(\ln(25x)\).

Answer:

\(\ln 25x + \ln 1\), \(\ln 25x\) (corresponding to the third and fourth options: \(\boldsymbol{\ln 25x + \ln 1}\), \(\boldsymbol{\ln 25x}\))