A savings account is started with an initial deposit of $500. The account earns 1.5% interest compounded annually.
Write an equation to represent the amount of money in the account as a function of time in years.
Find the amount of time it takes for the account balance to reach $800.

(a)A(t)=500((1+0.015)^(t))
(b)800=500((1+0.015)^(t))
800/500=(1+0.015)^(t)

t=31.567993958

(a)A(t)=500((1+0.015)^(t))
(b)800=500((1+0.015)^(t))
800/500=(1+0.015)^(t)

t=31.567993958

A) A=500(1+0.015)^t
b)800=500(1.015)^t
800/500=1.015^t
t=log(800/500)/log(1.015)
t=31.6 years

A) A=500(1+0.015)^t
b)800=500(1.015)^t
800/500=1.015^t
t=log(800/500)/log(1.015)
t=31.6 years

A)[tex]\bf ~~~~~~ \textit{compound interest earned amount} \\\\ a=p\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} a=\textit{accumulated amount}\\ p=\textit{original amount deposited}\to & \$500\\ r=rate\to 1.5\%\to \frac{1.5}{100}\to & 0.015\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\to & 1\\ t=years \end{cases} \\\\\\ a=500\left(1+\frac{0.015}{1}\right)^{1\cdot t}\implies a=500(1.015)^t[/tex]b)so, a = 800 then, thus[tex]\bf \textit{logarithm of exponentials}\\\\ log_a\left( x^ b \right)\implies b\cdot log_a(x)\\\\ \stackrel{a}{800}=500(1.015)^t\implies \cfrac{800}{500}=1.015^t\implies \cfrac{8}{5}=1.015^t \\\\\\ log\left( \cfrac{8}{5} \right)=log(1.015^t)\implies log\left( \cfrac{8}{5} \right)=t\cdot log(1.015) \\\\\\ \cfrac{log\left( \frac{8}{5} \right)}{log(1.015)}=t\implies 31.567993958\approx t[/tex]so, about 31 years, 6 months, 24 days and 11 and half hours.