Respuesta :
The formula for compounded interest is the following:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Where A is the final amount, P is the principal, the initial investment, r is the annual rate of interest, n is how many times it is compounded per year and t is the time in years.
So, assuming the given rate of interest is annual, we have:
[tex]\begin{gathered} A=25000 \\ P=10000 \\ r=3\%=0.03 \\ n=365 \\ t=? \end{gathered}[/tex]Where we got n = 365 because it is compounded daily and there are 365 days in an year.
So let's start by solving for t:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \frac{A}{P}=(1+\frac{r}{n})^{nt} \\ \log \frac{A}{P}=\log (1+\frac{r}{n})^{nt} \\ \log \frac{A}{P}=nt\log (1+\frac{r}{n}) \\ \frac{\log\frac{A}{P}}{n\log(1+\frac{r}{n})}=t \\ t=\frac{\log\frac{A}{P}}{n\log(1+\frac{r}{n})} \end{gathered}[/tex]Where the log base can be anyone, but it has to be the sme for both log.
Let's calculate the numerator and denominator separately first, using base 10:
[tex]\log _{}\frac{A}{P}=\log \frac{25000}{10000}=_{}\log 2.5=0.39794\ldots[/tex][tex]\begin{gathered} n\log (1+\frac{r}{n})=365\log (1+\frac{0.03}{365})=365\log (1+0.0000821918\ldots)= \\ =365\log (1.0000821918\ldots)=365\cdot0.000035694\ldots=0.013028\ldots \end{gathered}[/tex]Putting them together, we have:
[tex]t=\frac{\log\frac{A}{P}}{n\log(1+\frac{r}{n})}=\frac{0.39794\ldots}{0.013028\ldots.}=30.54\ldots\approx31[/tex]So, it will take between 30 and 31 years, closer to 31 years for it to grow to $25,000.
