Kimberly borrows 1000 dollars from Lucy,who charged interest of 5% per month ( which compounds monthly). What is the least integer number of months after which Kimberly will owe more than twice as much as she borrowed.

To find the least integer number of months after which Kimberly will owe more than twice as much as she borrowed, we can set up a formula using compound interest.

Let P be the principal amount borrowed (1000 dollars) and r be the monthly interest rate (5% or 0.05). The amount owed after n months can be calculated using the formula:

A = P(1 + r)^n

In this case, we want to find the least integer value of n for which the amount owed (A) is more than twice the principal (2P = 2 * 1000 = 2000 dollars).

So our equation becomes:

P(1 + r)^n > 2P

Dividing both sides of the equation by P:

(1 + r)^n > 2

Now, we need to solve for n. Take the logarithm (base 10 or natural logarithm) of both sides:

log(1 + r)^n > log(2)

Using the logarithmic property log(x^a) = a * log(x), this simplifies to:

n * log(1 + r) > log(2)

Finally, divide both sides of the equation by log(1 + r):

n > log(2) / log(1 + r)

Substituting the values of r = 0.05 (monthly interest rate) and log as natural logarithm:

n > ln(2) / ln(1 + 0.05)

Using a calculator, we can determine the approximate value of n:

n > 13.86

Since we are looking for the least integer value, the least number of months after which Kimberly will owe more than twice as much as she borrowed is 14 months.

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