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Question 1076332: How long will it take for an investment of $46,000 to grow to $48,600 if the investment earns interest at the rate of 10%/year compounded annually? (Round your answer to one decimal place.)
Answer by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website! The future value of an investment is P*(1+r/n)^nt, where P is the principal, r is the interest rate, n is the amount of compounding periods per year, and t is the number of years. So:
48600=46000*(1.10)^t
1.0565217391304347826086956521739=(1.10)^t
ln 1.0565217391304347826086956521739=ln (1.1)^t
ln 1.0565217391304347826086956521739=t ln(1.1)
1.0565217391304347826086956521739/ln 1.1=t
t=0.57687578105752534878401869924285, or .5769 years. ☺☺☺☺
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