$5,000.00 invested for a period of 6 years doubles to $10,000.00. What is the interest rate?
You will need to apply logarithms: <span>log(1 + rate) = {log(total) - log(Principal)} ÷ Years </span>log(1 + rate) = <span>{log(10,000) - log(5,000)} ÷ 6 </span>log(1 + rate) = (4 - 3.6989700043) / 6 log(1 + rate) =
<span>
<span>
<span>
0.301029957 / 6 </span></span></span>log(1 + rate) =
<span>
<span>
<span>
0.0501716595
</span>
</span>
</span>
Next, raise 10 to the power of
<span>
<span>
<span>
0.0501716595
</span>
</span>
</span>
which results in <span>
<span>
<span>
1.1224620317
</span>
</span>
</span>
This value represents 1 plus the interest rate, so the interest rate is 0.1224620317 or 12.24620317 percent. This concludes Part ONE. Now, onto Part TWO.
How many years does it take for $300 to increase to $9,600 at an annual rate of <span>12.24620317%? You will use the following formula: </span>(More logarithms involved). Years = {log(total) - log(Principal)} ÷ log(1 + rate) Years = {log(9,600) - log(300)} / log(<span>1.1224620317) </span>Years = (3.982271233 - 2.4771212547) / 0.050171659518 <span><span><span>Years = 1.5051499783 /
</span>
</span>
</span>.050171659518 Years = 30
