SOLUTION: Here's the problem: If there are originally 25 bacteria, at what rate are they growing if the population of the bacteria doubles in 5 hours? (Hint:solve 50=25(1+r)^5)

Algebra ->  Functions -> SOLUTION: Here's the problem: If there are originally 25 bacteria, at what rate are they growing if the population of the bacteria doubles in 5 hours? (Hint:solve 50=25(1+r)^5)      Log On


   

Question 893085: Here's the problem: If there are originally 25 bacteria, at what rate are they growing if the population of the bacteria doubles in 5 hours? (Hint:solve 50=25(1+r)^5)
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
You can solve the given equation, which seems to be based on hourly rate of increase r every hour.

log%28%2850%29%29=log%28%2825%29%29%2Blog%28%28r%5E5%29%29
log%28%2850%29%29=5log%28%281%2Br%29%29%2Blog%28%2825%29%29
5%2Alog%28%281%2Br%29%29=log%28%2850%29%29-log%28%2825%29%29
log%28%281%2Br%29%29=%281%2F5%29log%28%282%29%29
Evaluating the right side in base ten,
log%28%281%2Br%29%29=0.060206
10%5E0.060206=r%2B1=1.15

r=0.15, meaning rate is 15% each hour.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Here's the problem: If there are originally 25 bacteria, at what rate are they growing if the population of the bacteria doubles in 5 hours? (Hint:solve 50=25(1+r)^5)


50+=+25%281+%2B+r%29%5E5

2+=+%281+%2B+r%29%5E5 ------ Dividing both sides by 25

2%5E%281%2F5%29+=+%281+%2B+r%29%5E%285%281%2F5%29%29 ------ Raising each side to reciprocal of exponent, 5, or 1%2F5

1+%2B+r+=+2%5E%281%2F5%29

r+=+2%5E%281%2F5%29+-+1

r, or growth rate = .148698355, or rounded to: highlight_green%28highlight_green%28%0D%0A14.87%29%29%