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I'm going through applications of separable equations and came across an example of half-lives:

$$M(t)=\frac{M}{2}=Me^{-kt}$$

Factoring out $M$, $\frac{1}{2}=e^{-kt}$.

To solve for $t$, $\ln\left(\frac{1}{2}\right)=-kt$.

And then the answer is $t=\frac1k{\ln 2}$.

Why isn't the value for $t=-\frac1k{\ln \frac12}$?

Thanks:)

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    $\begingroup$ It is the same ! $\endgroup$
    – Claude Leibovici
    Aug 31, 2018 at 8:42
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    $\begingroup$ Recall the log sum rule: $\log(x) + \log(y) =\log(xy)$. Apply this with $y=1/x$ to derive $\log(x) = -\log(1/x)$. $\endgroup$
    – Winther
    Aug 31, 2018 at 8:44
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    $\begingroup$ Also, to complement @Winther's comment, note that $\log x^n = n\ \log x$ and $x^{-1}=\frac{1}{x}$. $\endgroup$
    – User123456789
    Aug 31, 2018 at 11:30

4 Answers 4

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It is the same since $$\ln(2)=-\ln(1/2).$$

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$\ln(1/2) = -kt$ implies $-\ln(2) = -kt$ implies $\dfrac{\ln(2)}{k}=t$ which is also $-\dfrac{\ln(1/2)}{k}=t$.

Reminder that $\ln(1/x) = -\ln(x)$.

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$\ln{\frac{1}{2}} = \ln{1} - \ln{2} = 0 - \ln{2} = -\ln{2}$

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Given the exponent rule for logarithms $\log(a^b) = b\log(a)$ with $a = 1/2$ and $b = -1$ we have

$$ \log(2) = \log((1/2)^{-1}) = -\log(1/2) $$

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