SOLUTION: A radioactive substance is produced from nuclear fallout. If 250g of this substance decays to 150g in 30years, what is the half-life of this substance? Solve algebraically usinglog

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Question 1206895: A radioactive substance is produced from nuclear fallout. If 250g of this substance decays to 150g in 30years, what is the half-life of this substance? Solve algebraically usinglogarithms. Answer to 2 decimal places.
Found 3 solutions by josgarithmetic, greenestamps, math_tutor2020:
Answer by josgarithmetic(39621) About Me  (Show Source):
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y=pe%5E%28-kt%29

If,...
150=250e%5E%28-k%2A30%29
ln%28150%29=ln%28250%29%2B%28-30k%29%2Aln%28e%29
-30k%2Bln%28250%29=ln%28150%29
-30k=ln%28150%29-ln%28250%29
k=%28ln%28250%29-ln%28150%29%29%2F30
k=%281%2F30%29%28ln%2825%2F15%29%29
k=%281%2F30%29%28ln%285%2F3%29%29
highlight_green%28k=0.017028%29

Half-Life?
1%2F2=1%2Ae%5E%28-xk%29
ln%281%2F2%29=ln%28e%5E%28-x%2A0.017028%29%29
-ln%282%29=-0.017028x%2Aln%28e%29
0.017028x%2A1=ln%282%29
highlight_green%28x=ln%282%29%2F0.017028%29
highlight%28x_halflife=40.7%29---------To 2 decimal places? 40.71 years

Answer by greenestamps(13203) About Me  (Show Source):
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Let x be the number half-lives in 30 years. Then, since 250g decays to 150g in 30 years,

150=250%280.5%29%5Ex
%280.5%29%5Ex=0.6
xlog%280.5%29=log%280.6%29
x=log%280.6%29%2Flog%280.5%29 = 0.7369656 to several decimal places

The half life is then

30%2F0.7369656 = 40.71 to 2 decimal places

ANSWER: 40.71 years

Answer by math_tutor2020(3817) About Me  (Show Source):
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Answer: 40.71 years

Explanation

One way to represent the half-life formula is:
y = A*(0.5)^(x/H)
where,
x = number of years
A = starting amount
H = half-life in years
y = final amount after x years
Optionally you can replace "years" with some other time unit.

Let's isolate H.
y = A*(0.5)^(x/H)
y/A = (0.5)^(x/H)
log( y/A ) = log( (0.5)^(x/H) )
log( y/A ) = (x/H)*log( 0.5 )
H*log( y/A ) = x*log( 0.5 )
H = x*log( 0.5 )/log( y/A )

Plug in x = 30, y = 150, A = 250
You should get the following
H = x*log( 0.5 )/log( y/A )
H = 30*log( 0.5 )/log( 150/250 )
H = 30*log( 0.5 )/log( 0.6 )
H = 40.707463465702 approximately
H = 40.71 years is the approximate half-life when rounding to 2 decimal places.