How long will it take for polonium-218 to decay to 30%?

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Homework Help Overview

The problem involves determining the time it takes for polonium-218 to decay to 30% of its original amount, using a decay function that includes an exponential term. The context is centered around radioactive decay and logarithmic equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the decay equation and the application of logarithmic properties to solve for time. There are attempts to clarify the manipulation of the equation and the role of the exponent.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to approach the logarithmic manipulation. There is an ongoing clarification regarding the structure of the equation and the correct application of logarithmic laws, but no consensus has been reached on the final steps.

Contextual Notes

One participant notes a lack of access to their textbook, which may be influencing their ability to recall necessary concepts. There is also a repeated emphasis on the division of the exponent by 3.1, indicating a potential point of confusion in the problem setup.

Edgar92
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1. The problem statement, all variables
a sample polonium-218 decays at A(t)=100(.5)^t/3.1
determine how many minutes it will take to decay to 30%

2. Homework Equations
log(a)/log(b)


3. The Attempt at a Solution

.3=100(.5)^t/3.1
.3/100=.5^t/3.1
 
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I realize this is simple but I am forgetting and don't have my textbook
 
You've already done this:

0.3=\frac{100(0.5)^t}{3.1}

\frac{0.3}{100}=\frac{(0.5)^t}{3.1}

So why don't you follow the same kind of procedure and multiply by 3.1 :smile:

and remember the log laws that if a^x=b then log_ab=x

and finally... (which you tried to show in the relevant equations but the equation was surprisingly cut short) ... log_cb=\frac{log_ab}{log_ac}
 
The exponent is divided by 3.1, not the equation
 
In that case, just use the log-exp switchy thingy immediately.
 
Edgar92 said:
The exponent is divided by 3.1, not the equation

Fine. But you can still take the log of both sides and get a linear equation in t, can't you?
 
Edgar92 said:
The exponent is divided by 3.1, not the equation

The log laws still apply.
 

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