Compound countinuously, it would only take blank years.
A=Pe^rt
A=6000
P=3000`
e=e
r=0.1
t=?
6000=300e^0.1*t
I meant to put 3000 for P
A=Pe^rt
6000/3000=3000/3000e(0.1t)
2=e^(0.1t)
In 2=In (e^(0.1t)
In 2= 0.1t(In e)
0.1t=In(2)
t=In(2)/0.1
=0.693147181/0.1
=6.931471806=7 years.
I agree, assuming the question was, "how long would it take for money to double at 10% compounded continuously?"
( In the good ol' days we had something called the rule of 72. Money doubles approximately if you multiply the time by the rate and you get 72.
So 72/10 = 7.2 years, which is close to the 7 years of your answer)
To find the value of t, we need to isolate it in the equation. Let's go step by step:
1. Start with the equation: 6000 = 300e^(0.1t). We want to solve for t.
2. Divide both sides of the equation by 300 to eliminate the coefficient:
6000/300 = e^(0.1t).
3. Simplify the left side of the equation:
20 = e^(0.1t).
4. Take the natural logarithm (ln) of both sides of the equation:
ln(20) = ln(e^(0.1t)).
5. Apply the logarithmic rule: ln(a^b) = b * ln(a).
ln(20) = 0.1t * ln(e).
Since ln(e) is equal to 1, we can simplify the equation further:
ln(20) = 0.1t.
6. Now, divide both sides of the equation by 0.1 to isolate t:
ln(20)/0.1 = t.
7. Use a calculator to evaluate the left side of the equation:
t ≈ 9.0254.
Therefore, approximately 9.0254 years would be required to reach $6000 when continuously compounded with a rate of 10% starting from $3000.