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# Logarithms question watch

1. Officials are testing athletes for doping at a sporting event. They model the concentration of a particular drug in an athlete's bloodstream using the equation D=6e^-1/10t, where D is the concentration of the drug in mg/l and t is the time in hours since the athlete took the drug.

Question:
It is impossible to detect this drug in the bloodstream if the concentration is lower than 3 mg/l. Show that this happens after t=-10ln(1/2).

The equation you have to solve is essentially 3=6e^-1/10t.

From there I got ln3=ln6e^-1/10t.
I rearranged it to get ln3/ln6=e^-1/10t.

Having seen the solution, I obviously approached it the wrong way, and I'm assuming my answer doesn't turn out right because of the step just mentioned. My question is why did my rearranging method not work?
2. (Original post by dont know it)
Officials are testing athletes for doping at a sporting event. They model the concentration of a particular drug in an athlete's bloodstream using the equation D=6e^-1/10t, where D is the concentration of the drug in mg/l and t is the time in hours since the athlete took the drug.

Question:
It is impossible to detect this drug in the bloodstream if the concentration is lower than 3 mg/l. Show that this happens after t=-10ln(1/2).

The equation you have to solve is essentially 3=6e^-1/10t.

From there I got ln3=ln6e^-1/10t.
I rearranged it to get ln3/ln6=e^-1/10t.

Having seen the solution, I obviously approached it the wrong way, and I'm assuming my answer doesn't turn out right because of the step just mentioned. My question is why did my rearranging method not work?
Couple of observations:

* In taking natural logs of both sides of the equation, you have to take the log of the whole left hand side and the whole right hand side. The right hand side is then ln(6e^-1/10t), which you can expand to ln(6) + ln(e^-1/10t), using ln(ab) = ln(a) + ln(b). But you could have saved yourself some bother here by dividing both sides by 6 before taking natural logs.

* Then you can simplify ln(e^-1/10t) by using ln(e^a) = a.

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