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# How long will it take you to triple your money if you invest it at a rate 6% compounded annually?

Last updated date: 17th Jul 2024
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Hint: Here we will simply use the formula of the compound interest. We will put the values of the variables in the formula and solve it to get the time that will be required to triple the principal amount. Compound interest is an interest that is calculated on the principal amount and the interest accumulated over the previous time period.

Formula used:
Formula of the compound interest, Final amount $= P{\left( {1 + \dfrac{R}{{100}}} \right)^T}$ where, P is the principal amount, R is the rate of interest and T is the time period.

Complete step by step solution:
The given rate of interest is $6\%$.
It is given that the amount is to be tripled in some span of time.
Let P be the principal amount. So the final amount i.e. triple amount will be equal to 3P.
Now we will use the formula of the compound interest and put the entire required variable in the equation. Therefore, we get
Final amount $= P{\left( {1 + \dfrac{R}{{100}}} \right)^T}$
Now we will put all the values i.e. final amount and rate of interest in the above equation. Therefore, we get
$\Rightarrow 3P = P{\left( {1 + \dfrac{6}{{100}}} \right)^T}$
Now we will solve this equation to get the value of the time period. Therefore, we get
$\Rightarrow 3 = {\left( {1 + \dfrac{6}{{100}}} \right)^T}$
Now we will apply log function to both side of the equation, we get
$\Rightarrow \log \left( 3 \right) = \log \left( {{{\left( {1 + \dfrac{6}{{100}}} \right)}^T}} \right)$
We know this property of the log function that $\log {a^b} = b\log a$. Therefore, we get
$\Rightarrow \log \left( 3 \right) = T \times \log \left( {1 + \dfrac{6}{{100}}} \right)$
$\Rightarrow \log \left( 3 \right) = T \times \log \left( {1.06} \right)$
$\Rightarrow T = \dfrac{{\log \left( 3 \right)}}{{\log \left( {1.06} \right)}}$
$\Rightarrow T = 18.85$
Simple interest $= \dfrac{{PRN}}{{100}}$ where, P is the principal amount, R is the rate of interest and N is the time period.