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I want to solve the next equation for the value of $t$ given the parameters

  • $R=7.5$
  • $\lambda=1.07\times10^{-2}$
  • Equation is $0.8R=R(1-e^{\lambda t})$

What I wonder, is that I was told this $t$ is 150, but what I do: $$0.8\frac{R}{R}=(1-e^{\lambda t})$$ $$0.8={(1-e^{\lambda t})}$$ $$(-0.2=-e^{\lambda t})\times-1$$ $$0.2=e^{\lambda t}$$ $$ln(0.2)=\lambda t$$ $$t=\frac{ln(0.2)}{\lambda}$$ $$t=\frac{−1.6094}{1.07\times10^{-2}}=−0.015041121$$ that is different from the 150. Which one is right?

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    I don't think you multiplied correctly in the final step. $\frac{−1.6094}{1.07\times10^{-2}}=−150.41121$ – 2017-02-28
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    dividing $10^{-2}$ is actually multiplying $10^{2}$, so the result should be greater than 100, and thus 150 might be the right answer – 2017-02-28
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    Whoa!! Thats right, thanks for say it. !! – 2017-02-28
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    @zahbaz is correct. As an aside, if you are unsure of whether an answer is correct, you should plug your final solution into the original equation to see if it works. In this case, you would see that your answer (and $150$) are incorrect and that $-150.41\ldots$ is the correct answer. – 2017-02-28
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    The final multiplication was wrong once I used the right exponent it was given the 150 result, what out for typos. But thanks for the the answers it turned to be very illustrative for the errors that can be made if one does not put attention in the work. @ Dr. Sonnhard Graubner gives the reason why this is positive instead of negative. Thanks for your time! – 2017-02-28

2 Answers 2

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after dividing by $r$ we get $$\frac{4}{5}=1-e^{-\lambda t}$$ from here we get $$e^{-\lambda t}=\frac{1}{5}$$ taking the logarithm on both sides we get $$-\lambda t=\ln\left(\frac{1}{5}\right)$$ thus we have $$t=\frac{\ln\left(\frac{1}{5}\right)}{-\lambda}$$ plugging the values as given in this equation we get $$t\approx 150.4145$$

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I'm guessing you entered $\ln(0.2)/{1.07\times10^{-2}}$ on a calculator.

What you meant to enter was $\ln(0.2)/\left({1.07\times10^{-2}}\right)$.