Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago

# exponetial?

how to solve this problems

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• 6 months ago

I like to use this equation for all exponential growth and decay questions:

a(t) = a e^(kt)

Where "a" is the initial amount (we can use 100 here)

e is the mathematical constant that is approximately equal to 2.71828

t is the amount of time (in this case years)

k is the decay (or grown if positive) constant.

Looking at the graph, the half-life is when a(t) = 50. This is at t = 25. So knowing this information we can derive our equation to find any point:

a(t) = a e^(kt)

a(25) = 100 e^(k * 25)

50 = 100 e^(25k)

0.5 = e^(25k)

ln(0.5) = 25k

k = ln(0.5) / 25

The log of a number less than 1 is negative, making this a decay function.

Now our equation is:

a(t) = a e^(kt)

a(t) = 100 e^(t * ln(0.5) / 25)

So what percentage exists after 45 years? Since it's slightly less than 2 full half-lives, we should expect to have a value near 25% (the result of two half-lives). Substitute 45 in for t and simplify:

a(45) = 100 e^(45 * ln(0.5) / 25)

a(45) = 100 e^(-1.24766)

(I rounded here since I had to, but my calculator still uses the non-rounded value to reduce the chance of errors due to rounding)

a(45) = 100(0.28717)

a(45) = 100(0.28717)

a(45) = 28.717

So after 45 years, there is still 28.717% (rounded to 3DP) remaining of the initial compound.

• Mike G
Lv 7
6 months ago

i) From the graph

50% after 25 years

Half-life = 25 years

ii) A = 100(0.5)^(45/25)

= 28.717%

• 6 months ago

After how many years is the amount halved?

p = 100 * 2 ^(-t/25)

Plug in 45