You want to make an investment in a continuously compounding account over a period of ten years. What interest rate is required for your investment to double in that time period? Round the logarithm value to the nearest hundredth and the answer to the nearest tenth.

-I don't know what equation to use. Any help will be apprecited. Please show steps so I know what to do for future problems. Thank you

you want e^10r = 2

10r = ln2 = 0.693
r = 0.0693

so, a rate of 6.93% will do the trick

To find the interest rate required for your investment to double in a continuously compounding account over a period of ten years, you can use the formula for compound interest:

A = P * e^(rt),

Where:
A is the amount after time t,
P is the principal investment,
e is the mathematical constant approximately equal to 2.71828,
r is the interest rate (in decimal form), and
t is the time period.

In this case, since you want the investment to double, the amount A will be twice the initial principal investment P. So, we can rewrite the formula as:

2P = P * e^(rt).

Now, we can simplify and solve for the interest rate r:

2 = e^(rt).

To solve for r, we can take the natural logarithm (ln) of both sides:

ln(2) = ln(e^(rt)).

Using the property of logarithms, we can bring the exponent rt down:

ln(2) = rt * ln(e).

Since ln(e) is equal to 1, we can simplify further:

ln(2) = rt.

Now, we can solve for r by dividing both sides of the equation by t:

r = ln(2) / t.

To calculate the interest rate for your investment to double in ten years, substitute the values into the equation:

r = ln(2) / 10.

Using a calculator, find the natural logarithm of 2, which is approximately 0.693. Then divide by 10:

r ≈ 0.693 / 10 ≈ 0.0693.

Rounding to the nearest hundredth, the interest rate required for your investment to double in ten years is approximately 0.07 (or 7% when expressed as a percentage).

I hope this helps you understand the process of finding the interest rate using continuous compounding. Let me know if you need further assistance!

To find the interest rate required for your investment to double in a continuously compounding account over a period of ten years, you can use the formula for compound interest:

A = P * e^(r * t)

Where:
A is the final amount (double the initial investment)
P is the principal amount (initial investment)
e is Euler's number, approximately 2.71828
r is the interest rate (annual percentage rate)
t is the time duration in years

Let's set up the equation using the given information:

2P = P * e^(r * 10)

To solve for the interest rate (r), divide both sides of the equation by P:

2 = e^(r * 10)

Take the natural logarithm (ln) of both sides to isolate the exponent:

ln(2) = r * 10 * ln(e)

Since ln(e) is equal to 1, the equation simplifies to:

ln(2) = 10r

Now, divide both sides by 10 to solve for r:

r = ln(2) / 10

Using a calculator, evaluate ln(2) and divide it by 10:

r ≈ 0.0693

Rounding the logarithm value to the nearest hundredth and the answer to the nearest tenth, the interest rate required for your investment to double in ten years is approximately 6.9%.