please help me with the following:
The value of a particular investment follows a pattern of exponential growth. In the year 2000, you invested money in a money market account. The value of your investment t years after 2000 is given by the exponential growth model a=3000e^0.053t When will the account be worth $5097?
I got 10 years, but my answer choices are: a. 2010, b.2011, c.2009, d. 2012.
And I don't understand how you took the natural log of both sides?
You add t to the year 2000 in this case, so the answer is a.
Taking the natural log (to base e) was in my case done with a hand calculator. The natural log of e^x is x.
To find out when the account will be worth $5097, we need to solve the equation a = 5097 for t. Given that the value of the investment t years after 2000 is given by the exponential growth model a = 3000e^(0.053t), we can substitute a = 5097 into the equation and solve for t.
So, let's set up the equation:
5097 = 3000e^(0.053t)
To solve this equation, we can follow these steps:
Step 1: Divide both sides of the equation by 3000 to isolate e^(0.053t).
(5097/3000) = e^(0.053t)
Step 2: Take the natural logarithm (ln) of both sides.
ln(5097/3000) = ln(e^(0.053t))
Step 3: Use the property of logarithms that states ln(e^x) = x to simplify the equation.
ln(5097/3000) = 0.053t
Step 4: Divide both sides of the equation by 0.053 to solve for t.
t = ln(5097/3000) / 0.053
Using a calculator, you can find the value of ln(5097/3000), which is approximately 0.476. Dividing 0.476 by 0.053 gives:
t ≈ 0.476 / 0.053
t ≈ 8.98
Therefore, the account will be worth $5097 approximately 8.98 years after the year 2000.
Set 3000e^0.053t = $5097
e^0.053t = 5097/3000 = 1.699
Take the natural log of both sides
.053 t = 0.53
t = _?_ years