If monthly payments p are deposited in a savings account paying an annual interest rate r, then the amount A in the account after n years is given by
A= p(1+r/12)[(1+r/12)^12n - 1]/ (r/12)
Graph A for the values of p and r.
p = 100, r = 0.06
Estimate n for A = $200,000. (Round your answer to one decimal place.)
r/12 = .06/12 = .005
1+r/12 = 1.005
A = 100 (1.005) [ 1.005^12n -1]/.005
200,000 = 20,000 (1.005)[ 1.005^12n -1]
2000 = [ 1.005^12n -1]
2001 = 1.005^12n
log 2001 = 12 n log 1.005
12 n = 1524.08
n = 127 years
check my arithmetic !
To estimate the value of n for A = $200,000, we can rearrange the equation for A and solve for n.
The equation for A is:
A = p(1 + r/12)[(1 + r/12)^(12n) - 1]/(r/12)
We know the values for p = 100 and r = 0.06. Let's substitute these values into the equation:
A = 100(1 + 0.06/12)[(1 + 0.06/12)^(12n) - 1]/(0.06/12)
Simplifying the equation further:
A = 100(1 + 0.005)[(1 + 0.005)^(12n) - 1]/0.005
A = 100(1.005)[(1.005)^(12n) - 1]/0.005
Now, let's substitute A = $200,000 into the equation:
$200,000 = 100(1.005)[(1.005)^(12n) - 1]/0.005
To solve for n, we need to isolate the term with the exponent:
$200,000 * 0.005 = 100(1.005)[(1.005)^(12n) - 1]
$1,000 = (1.005)[(1.005)^(12n) - 1]
Now, divide both sides by 1,000:
(1.005)[(1.005)^(12n) - 1] = 1
Next, divide both sides by 1.005:
[(1.005)^(12n) - 1] = 1/1.005
Now, simplify the right side of the equation:
[(1.005)^(12n)] - 1 = 0.99502487562
To solve for n, we need to isolate the term with the exponent again. Let's add 1 to both sides:
(1.005)^(12n) = 0.99502487562 + 1
(1.005)^(12n) = 1.99502487562
Now, take the logarithm of both sides using the base 1.005:
log base 1.005 [(1.005)^(12n)] = log base 1.005 (1.99502487562)
12n = log base 1.005 (1.99502487562)
Finally, divide both sides by 12 to solve for n:
n = log base 1.005 (1.99502487562) / 12
Calculating this expression will give us an estimate of n for A = $200,000.