Question

Zeke deposited $300 into a savings account. The interest rate on the account is 3%, compounded continuously. To determine how long it will take to triple his money, Zeke wrote the following equation, where t represents time, in years.

900=300e^0.03t
Which equation is equivalent to the one Zeke wrote?
ln900=ln(9t)
ln900=9t
ln3=0.03t
ln3=ln(0.03t)

Answer 1

`\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( (x)/(y)\right)\implies \log_a(x)-\log_a(y) \end{array}`

`\bf \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\textit{let's use this one}}{log_a a^x = x}\qquad \quad a^(log_a x)=x \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 900=300e^(0.03t)\implies \log_e(900)=\log_e(300e^(0.03t))`

`\bf \log_e(900)=\log_e(300)+\log_e(e^(0.03t))\implies \ln(900)=\ln(300)+0.03t\cdot \ln(e) \\\\\\ \ln(900)-\ln(300)=0.03t\implies \ln\left( \cfrac{900}{300} \right)=0.03t\implies \ln(3)=0.03t`