If you open a savings account that earns 7.5% simple interest per year, what is the minimum number of years you must wait to double your balance? Suppose you open another account that earns 7% interest compounded yearly. How many years will it take now to double your balance?

Respuesta :

Answer:

Case 1 = 13.33 years

Case 2 = 10.25 years

Explanation:

Case 1

Interest rate = 7.5% = 0.075

Future value = 2 × Present value

Now,

Future value = Present value × [ 1 +  ( Rate × Time ) ]

or

2 × Present value = Present value × [ 1 +  ( 0.075 × Time ) ]

or

2 = 1 +  ( 0.075 × Time )

or

1 = 0.075 × Time

or

Time = 13.33 years

Case 2

Interest rate, r = 7% = 0.07

Future value = 2 × Present value

Now,

Future value = Present value × ( 1 + r )ⁿ

here,

n is the time

thus,

2 × Present value = Present value × ( 1 + 0.07 )ⁿ

or

2 = 1.07ⁿ

taking log both the sides

log(2) = log(1.07ⁿ)

also,

log(aᵇ) = b × log(a)

thus,

log(2) = n × log(1.07)

or

0.301 = n × 0.02938

or

n = 10.25 years

In Case 1, the minimum number of years to double the balance is 13.33 years, whereas Case 2 needs 10.25 years.

Step-by-step Computation

Case 1

[tex]\text{Interest rate} = 7.5[/tex]% or [tex]0.075[/tex]

[tex]\text{Future value} = 2[/tex] × [tex]\text{Present value}[/tex]

Now,

[tex]\text{Future value = Present value}[/tex] ×[tex][ 1 + ( \text{Rate}[/tex] ×[tex]\text{Time} ) ][/tex]

or

[tex]2[/tex] × [tex]\text{Present value = Present value}[/tex] ×[tex][ 1 + ( 0.075[/tex] × [tex]\text{Time ) ]}[/tex]

or

[tex]2 = 1 + ( 0.075[/tex] × [tex]\text{Time )}[/tex]

or

[tex]1 = 0.075[/tex] × [tex]\text{Time}[/tex]

or

[tex]\text{Time = 13.33 years}[/tex]

Case 2

[tex]\text{Interest rate, r = 7}[/tex]% [tex]= 0.07[/tex]

[tex]\text{Future value = 2}[/tex] × [tex]\text{Present value}[/tex]

Now,

[tex]\text{Future value = Present value}[/tex] × [tex](1+r)^{n}[/tex]

here,

[tex]\text{n = Time}[/tex]

thus,

[tex]2[/tex] × [tex]\text{Present value = Present value}[/tex] × [tex]( 1 + 0.07 )^{n}[/tex]

or

[tex]2 = 1.07^{n}[/tex]

[tex]\text{Taking log both the sides}[/tex]

[tex]\text{log(2)} = \text{log}(1.07^{n})[/tex]

also,

[tex]\text{log}(a^{b}) = b[/tex] × [tex]\text{log(a)}[/tex]

thus,

[tex]\text{log}(2) = n[/tex] × [tex]\text{log}(1.07)[/tex]

or

[tex]0.301 = n[/tex] × [tex]0.02938[/tex]

or

[tex]n = 10.25 \text{years}[/tex]

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