alyssa deposits $100 every month into an account earning an annual interest rate of 3%, compounded monthly…

alyssa deposits $100 every month into an account earning an annual interest rate of 3%, compounded monthly. how many months would it be until alyssa had $2,000 in the account, to the nearest month? use the following formula to determine your answer.\n\n$a = d\\left(\\frac{(1 + i)^n - 1}{i}\\right)$\n\n$a=$ the future value of the account after $n$ periods\n$d=$ the amount invested at the end of each period\n$i=$ the interest rate per period\n$n=$ the number of periods

alyssa deposits $100 every month into an account earning an annual interest rate of 3%, compounded monthly. how many months would it be until alyssa had $2,000 in the account, to the nearest month? use the following formula to determine your answer.\n\n$a = d\\left(\\frac{(1 + i)^n - 1}{i}\\right)$\n\n$a=$ the future value of the account after $n$ periods\n$d=$ the amount invested at the end of each period\n$i=$ the interest rate per period\n$n=$ the number of periods

Answer

Explicación:

Paso 1: Identificar variables y calcular la tasa de interés periódica.

El valor futuro deseado es $A = $2,000$. El depósito mensual es $d = $100$. La tasa de interés anual es $3%$, o $0.03$. Dado que el interés se capitaliza mensualmente, la tasa de interés por período (mes) es $i = \frac{0.03}{12} = 0.0025$.

Paso 2: Sustituir los valores en la fórmula dada.

La fórmula para el valor futuro de una anualidad es $A = d \left( \frac{(1+i)^n - 1}{i} \right)$. Sustituyendo los valores conocidos: $$2000 = 100 \left( \frac{(1+0.0025)^n - 1}{0.0025} \right)$$

Paso 3: Simplificar la ecuación.

Dividimos ambos lados por 100: $$ \frac{2000}{100} = \frac{(1.0025)^n - 1}{0.0025} $$ $$ 20 = \frac{(1.0025)^n - 1}{0.0025} $$ Multiplicamos ambos lados por 0.0025: $$ 20 \times 0.0025 = (1.0025)^n - 1 $$ $$ 0.05 = (1.0025)^n - 1 $$ Sumamos 1 a ambos lados: $$ 0.05 + 1 = (1.0025)^n $$ $$ 1.05 = (1.0025)^n $$

Paso 4: Resolver para n usando logaritmos.

Tomamos el logaritmo natural (ln) en ambos lados de la ecuación: $$ \ln(1.05) = \ln((1.0025)^n) $$ Usando la propiedad del logaritmo $\ln(x^y) = y \ln(x)$: $$ \ln(1.05) = n \ln(1.0025) $$ Despejamos n: $$ n = \frac{\ln(1.05)}{\ln(1.0025)} $$

Paso 5: Calcular el valor de n y redondear.

Calculamos los valores de los logaritmos: $$ \ln(1.05) \approx 0.04879016 $$ $$ \ln(1.0025) \approx 0.00249688 $$ Ahora calculamos n: $$ n \approx \frac{0.04879016}{0.00249688} \approx 19.53999 $$ Redondeando al mes más cercano, $n \approx 20$.

Respuesta:

20 meses