00:01
In this question we have to find the time in years in which an investment of $10 ,000 grows to $35 ,000.
00:15
Also we are given certain information like the principal p is given to be $10 ,000.
00:26
The amount a is $35 ,000.
00:38
And the rate of interest are is 4%.
00:50
Now, let us suppose that n represents the time or we can say the number of years that we are supposed to calculate.
01:09
Now it is known that for the situation where the rate of interest is compounded continuously, the amount a equals to p multiplied by e to the power r multiplied by n.
01:31
Now simply substituting the values of these terms we get 35 ,000 equals to 10 ,000 multiplied by e to the power 0 .04 as the rate of interest is 4 percent so 4 divided by 100.
01:55
Is 0 .04 multiplied by n.
02:00
Now 35 ,000 divided by 10 ,000 equals to e to the power 0 .04n.
02:17
That means 3 .5 equals to e to the power 0 .04n...