- #1

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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

- Thread starter Edgar92
- Start date

- #1

- 24

- 0

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

- #2

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- 0

I realize this is simple but I am forgetting and don't have my textbook

- #3

Mentallic

Homework Helper

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[tex]0.3=\frac{100(0.5)^t}{3.1}[/tex]

[tex]\frac{0.3}{100}=\frac{(0.5)^t}{3.1}[/tex]

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

and remember the log laws that if [tex]a^x=b[/tex] then [tex]log_ab=x[/tex]

and finally... (which you tried to show in the relevant equations but the equation was surprisingly cut short) ... [tex]log_cb=\frac{log_ab}{log_ac}[/tex]

- #4

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The exponent is divided by 3.1, not the equation

- #5

Char. Limit

Gold Member

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In that case, just use the log-exp switchy thingy immediately.

- #6

Dick

Science Advisor

Homework Helper

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

- #7

Mentallic

Homework Helper

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The log laws still apply.The exponent is divided by 3.1, not the equation

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