In what time will the principal double itself at the same rate of interest as it would be 3/5 of the amount in 10 years?
To solve this problem, we can use the concept of compound interest. When the principal doubles, it means the amount is twice the principal. Let's denote the time it takes for the principal to double as 't' years.
According to the given information, the amount in 10 years would be 3/5 of the amount when the principal doubles. Mathematically, we can express this as:
Principal * (1 + r/100)^(10) = (3/5) * Principal * (1 + r/100)^t
Where 'r' represents the rate of interest.
We can cancel out the common factors on both sides:
(1 + r/100)^(10) = 3/5 * (1 + r/100)^t
To find 't', we can take the logarithm of both sides. Let's assume a base of 10 for simplicity:
log((1 + r/100)^(10)) = log((3/5) * (1 + r/100)^t)
Using the logarithmic property log(a^b) = b * log(a), we can rewrite the equation as:
10 * log(1 + r/100) = log((3/5)) + t * log(1 + r/100)
Now, we can isolate 't' by rearranging the equation:
t * log(1 + r/100) = 10 * log(1 + r/100) - log((3/5))
t = (10 * log(1 + r/100) - log((3/5))) / log(1 + r/100)
Using the appropriate interest rate and performing the calculations, we can determine the value of 't' in years.