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solve: \\ln 2x + \\ln 2 = 0\ x = \\boxed{\\space}\ \ solve: e^{2x + 5} …

Question

solve: \ln 2x + \ln 2 = 0\
x = \boxed{\space}\
\
solve: e^{2x + 5} = 4\
\bigcirc\\ x = -\frac{5}{2}\
\bigcirc\\ x = \frac{\ln 4}{2} - 5\
\bigcirc\\ x = \frac{\ln 4 - 5}{2}

Explanation:

Step1: Use logarithm property

We know that \(\ln a+\ln b = \ln(ab)\), so \(\ln 2x+\ln 2=\ln(2x\times2)=\ln(4x)\). The equation becomes \(\ln(4x) = 0\).

Step2: Convert to exponential form

Since \(\ln y = z\) is equivalent to \(y = e^{z}\), for \(\ln(4x)=0\), we have \(4x=e^{0}\).

Step3: Solve for x

We know that \(e^{0} = 1\), so \(4x = 1\), then \(x=\frac{1}{4}\).

Answer:

\(\frac{1}{4}\)