Question

How long will it take to earn \[\$ 7000\] on a \[\$ 3500\] investment at a rate of \[7.8\% \] compounding continuously?

Answer 1

Hint: Here we need to find the time in years. In the question let us assume that it is compounded per year. We know the formula of compound interest \[A = P{(1 + r)^n}\] , where ‘A’ is amount, ‘P’ is principal amount, ‘r’ is rate of interest per year and ‘n’ is the time in year.

Complete step by step solution:
Given,
 \[A = 7000\] , \[P = 3500\] and \[r = 7.8\% \] .
We know that \[A = P{(1 + r)^n}\] .
Substituting we have,
 \[7000 = 3500{\left( {1 + \dfrac{{7.8}}{{100}}} \right)^n}\] .
Divide the whole equation by 3500
 \[\dfrac{{7000}}{{3500}} = {\left( {1 + \dfrac{{7.8}}{{100}}} \right)^n}\]
 \[2 = {\left( {1 + \dfrac{{7.8}}{{100}}} \right)^n}\]
 \[{\left( {1 + \dfrac{{7.8}}{{100}}} \right)^n} = 2\]
Taking logarithm on both sides we have,
 \[\log {\left( {1 + 0.078} \right)^n} = \log 2\]
Applying logarithm that is \[\log {\left( m \right)^n} = n\log \left( m \right)\] .
 \[n.\log \left( {1.078} \right) = \log 2\]
 \[n = \dfrac{{\log 2}}{{\log \left( {1.078} \right)}}\]
We know \[\log (1.078) = 0.03261\] and \[\log 2 = 0.3010\]
 \[n = \dfrac{{0.3010}}{{0.03261}}\]
 \[ \Rightarrow n = 9.23\] .
That is 9.23 years.
So, the correct answer is “9.23 years.”.

Note: Simple interest and compound interest are different. The simple interest is a type of interest that is applied to the amount borrowed or invested for the entire duration of the loan, without taking any other factors into account, such as past interest or any other financial considerations. The compound interest is the interest which is calculated on the principal and the interest that is accumulated over as ‘interest on interest’.