situation: holly wants to save money for an emergency. holly invests $1,100 in an account that pays an…

situation: holly wants to save money for an emergency. holly invests $1,100 in an account that pays an interest rate of 8.75%. how many years will it take for the account to reach $6,400? round your answer to the nearest hundredth. enter the correct answer. a = p(1 + r)^t p = amount of money invested r = interest rate percentage in decimal form no. substitute known values into the equation. then take the log of each side. rewrite the equation using the property: loga u^n = n loga u. try again.

situation: holly wants to save money for an emergency. holly invests $1,100 in an account that pays an interest rate of 8.75%. how many years will it take for the account to reach $6,400? round your answer to the nearest hundredth. enter the correct answer. a = p(1 + r)^t p = amount of money invested r = interest rate percentage in decimal form no. substitute known values into the equation. then take the log of each side. rewrite the equation using the property: loga u^n = n loga u. try again.

Answer

Explanation:

Step1: Identify values

$A = 6400$, $P=1100$, $r = 0.0875$ Substitute into $A = P(1 + r)^t$: $6400=1100(1 + 0.0875)^t$

Step2: Simplify the equation

Divide both sides by 1100: $\frac{6400}{1100}=(1.0875)^t$, so $\frac{64}{11}=(1.0875)^t$

Step3: Take the natural - log of both sides

$\ln(\frac{64}{11})=\ln(1.0875^t)$

Step4: Use the logarithm property

$\ln(\frac{64}{11}) = t\ln(1.0875)$

Step5: Solve for t

$t=\frac{\ln(\frac{64}{11})}{\ln(1.0875)}$ $\ln(\frac{64}{11})\approx\ln(64)-\ln(11)\approx4.15888 - 2.3979\approx1.76098$ $\ln(1.0875)\approx0.084$ $t=\frac{1.76098}{0.084}\approx20.96$

Answer:

$20.96$