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The math problem:

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And this is my solution:

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And the book's solution is:

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I tried using a math website (Mathway) to see what solution they would get and they got the same as mine:

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Is the book incorrect?

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    $\begingroup$ $\ln(1/2)=-\ln(2)$ $\endgroup$ Oct 9, 2020 at 23:01

1 Answer 1

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The book is not incorrect, it's the same answer just expressed differently. The quotient rule for logarithms states that $\log_{b}\left(\frac{x}{y}\right) = \log_{b}(x) - \log_{b}(y)$. If $x = 1$ as it does in $\ln\left(\frac{1}{2}\right)$ then the quotient rule would tell you that $\ln\left(\frac{1}{2}\right) = -\ln(2)$ as $\ln(1) = 0$. Thus, $\frac{1}{2}\ln\left(\frac{1}{2}\right) = -\frac{1}{2}\ln\left(2\right)$.

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  • $\begingroup$ I see what you mean. Thank you for clarifying this. It went over my head when I was solving the equation. It now makes sense. $\endgroup$
    – coder3
    Oct 11, 2020 at 3:04

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