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Exercise 6.6 · Q89

Q.The rate of decay of a certain substance is directly proportional to the amount present at that instant. Initially there are 25 gms of the substance, and two hours later it is found that 9 gms are left. Find the amount left after one more hour.

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Let x=25e−ktx=25e^{-kt}. At t=2t=2, x=9x=9: 9=25e−2k⇒e−2k=925⇒e−k=359=25e^{-2k}\Rightarrow e^{-2k}=\dfrac{9}{25}\Rightarrow e^{-k}=\dfrac35 (taking the positive root). At t=3t=3: $x=25e^{-3k}=25(e^{-k})^3=25\left(\dfrac35\ …

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