Jayden invested $22,000 in an account paying an interest rate of 2.5% compounded continuously. Assuming no deposits or withdrawals are made, how long would it
take, to the nearest tenth of a year, for the value of the account to reach $30,400?
PLEASE HELP

12.9

Step-by-step explanation:

30400=

30400=

\,\,22000e^{0.025t}

22000e

0.025t

Plug in

\frac{30400}{22000}=

22000

30400

​

=

\,\,\frac{22000e^{0.025t}}{22000}

22000

22000e

0.025t

​

Divide by 22000

1.3818182=

1.3818182=

\,\,e^{0.025t}

e

0.025t

\ln\left(1.3818182\right)=

ln(1.3818182)=

\,\,\ln\left(e^{0.025t}\right)

ln(e

0.025t

)

Take the natural log of both sides

\ln\left(1.3818182\right)=

ln(1.3818182)=

\,\,0.025t

0.025t

ln cancels the e

\frac{\ln\left(1.3818182\right)}{0.025}=

0.025

ln(1.3818182)

​

=

\,\,\frac{0.025t}{0.025}

0.025

0.025t

​

Divide by 0.025

12.9360062=

12.9360062=t

t = 12.9

12.9

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.