Algebra help?

Question

Can someone help me figure this out. I tried different formulas, but I am not sure which is correct.

A woman deposits $50,000 in a savings account with 4% continuously compounded interest. How many years must she wait until the balance has doubled?

Answer 1

whats in it for me?

Answer 2

Using equation: (P) x (1 + 0.04)^n = B
Where P = Starting Principle = $50,000
Where n = Time Period in Years
Where B = Ending Balance = $100,000
If this account compounds once per year @ 4%, then
when: n = 17.67 years will yield B = $99,988.28
when: n = 17.68 years will yield B = $100,027.51
Answer: Approximately 17.675 years

Answer 3

For continuous compounding there is an exponential solution:

The final balance is $100,000

100,000= 50,000 e^(it)

Where i is interest rate and t is number of years.

2= e^(0.04t)

Therefore

Ln(2)= Ln (e^(0.04t))

Ln(2) = 0.04t

t = 0.04/0.693

t = 17.3 yrs
.

Answer 4

Let N be the number of the years when the Amount doubles from $50,000 to $1,00,000.
The formula states:
A=P(1+0.04)^N
So after N years,
=>1,00,000=50,000(1.04)^N
=>2=(1.04)^N
=>log2=N[log(1.04)]
=>N=loge2/loge(1.04)
=>N=17.7 years

Answer 5

amount at end is twice than started ... so

2 = e^(.04 *T) take LN

ln(2) = .04T
ln(2) / .04 = t