If you want to double your investment of $1,200 using a savings account with a 6% annual interest rate compounded annually, how many years will it take?

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To determine how many years it will take to double an investment of $1,200 with a 6% annual interest rate compounded annually, we can use the formula for compound interest:

$$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$

Where:
- $$ A $$ is the amount of money accumulated after n years, including interest.
- $$ P $$ is the principal amount (initial investment).
- $$ r $$ is the annual interest rate (in decimal form).
- $$ n $$ is the number of times that interest is compounded per year.
- $$ t $$ is the time the money is invested for in years.

Given:
- $$ P = \$1,200 $$
- $$ r = 6\% = 0.06 $$
- $$ n = 1 $$ (compounded annually)
- $$ A = \$2,400 $$ (double the initial investment)

We need to solve for $$ t $$ when $$ A = \$2,400 $$.
Solve the equation by following steps:
- step0: Solve for $$t$$:
  $$1200\left(1+0.06\right)^{t}=2400$$
- step1: Simplify:
  $$1200\left(\frac{53}{50}\right)^{t}=2400$$
- step2: Multiply by the reciprocal:
  $$1200\left(\frac{53}{50}\right)^{t}\times \frac{1}{1200}=2400\times \frac{1}{1200}$$
- step3: Multiply:
  $$\left(\frac{53}{50}\right)^{t}=2$$
- step4: Take the logarithm of both sides:
  $$\log_{\frac{53}{50}}{\left(\left(\frac{53}{50}\right)^{t}\right)}=\log_{\frac{53}{50}}{\left(2\right)}$$
- step5: Evaluate the logarithm:
  $$t=\log_{\frac{53}{50}}{\left(2\right)}$$
The number of years it will take to double the investment of $1,200 with a 6% annual interest rate compounded annually is $$ t = \log_{\frac{53}{50}}{(2)} $$.
It will take approximately 12 years to double the investment of $1,200 with a 6% annual interest rate compounded annually.

If you want to double your investment of $1,200 using a savings account with a 6% annual interest rate compounded annually, how many years will it take?

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