1. how long will it take for an investment to triple if it is compounded annually at 5%? (round answer to the nearest tenth in years.) I got 22.0 years but not sure if it's right.
2. suppose that in a certain area the consumption of electricity has increased at a continuous rate of 6%. if it continued to increase at this rate, find the number of years before four times as much electricity would be needed.
3. the rate of inflation is such that the price of a car costing $17,000 will double in ten years. what is the annual rate of inflation? what will the price of this car be in 7 years?
2. suppose that in a certain area the consumption of electricity has increased at a continuous rate of 6%. if it continued to increase at this rate, find the number of years before four times as much electricity would be needed.
3. the rate of inflation is such that the price of a car costing $17,000 will double in ten years. what is the annual rate of inflation? what will the price of this car be in 7 years?
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1)
1.05^n = 3 where n is the number of years.
so n*log(1.05) = log(3)
n = log(3) / log(1.05)
n = 22.517 = 22.5 years.
As the interest is compounded annually, the investment will not attain 3 times its initial value until the 23rd interest payment.
2)
n = log(4) / log(1.06)
n = 23.8 years
3)
If i is the annual rate of inflation and it takes 10 years for the cost to double
(1 + i)^10 = 2
10*log(1 + i) = log(2)
log (1 + i) = log(2) / 10 = 0.030103
l + i = antilog (0.030103) = 1.07177
i = 7.2%
Now 7 * log (1 + i) = log (C / 17000)
0.7 * log (2) = log (C / 17000) = 0.21072
C = 17000 * antilog (0.21072) = $27616.58 after 7 years
1.05^n = 3 where n is the number of years.
so n*log(1.05) = log(3)
n = log(3) / log(1.05)
n = 22.517 = 22.5 years.
As the interest is compounded annually, the investment will not attain 3 times its initial value until the 23rd interest payment.
2)
n = log(4) / log(1.06)
n = 23.8 years
3)
If i is the annual rate of inflation and it takes 10 years for the cost to double
(1 + i)^10 = 2
10*log(1 + i) = log(2)
log (1 + i) = log(2) / 10 = 0.030103
l + i = antilog (0.030103) = 1.07177
i = 7.2%
Now 7 * log (1 + i) = log (C / 17000)
0.7 * log (2) = log (C / 17000) = 0.21072
C = 17000 * antilog (0.21072) = $27616.58 after 7 years