Question

The amount of money invested in a certain account increases according to the following function, where

y0
is the initial amount of the investment, and
y
is the amount present at the time
t
(in years).
y=yoe^0.0525t
After how many years will the initial investment be doubled? Do not round any intermediate computations, and round your answer to the nearest tenth.

Answer 1

approximately 13.2 years

Explanation:

Interpreting The Question

So we start off with the general equation:
`y=y_0 * e^(0.0525t)`, in this exponential function, the
`y_0` represents the initial value or the y-intercept, and when this value becomes doubled that means it's being multiplied by 2. Conveniently this value is being multiplied by:
`e^(0.0525t)`, meaning this has to be equal to two for the initial investment to double.

Logarithms

Logarithms are going to be super useful in these types of questions, and are essentially the inverse of an exponential.

So if I have a log in the form:
`log_ba=c` that means that "I have to raise 'b' to the power of 'c' to get the result 'a'" or in equation form:
`log_ba=c\implies b^c=a`

Generally we define the base 'b' to some value, but there are two logarithms that show up a lot: log which has a base of 10 and ln pronounced "natural log" which has a base of "e"

This just means that:
`log(20) \implies log_(10)(20)` even if the base wasn't explicitly written

Likewise:
`ln(20)\implies log_(e)(20)` even if the base wasn't explicitly written.

Solving for T

we know that

`e^(0.0525t) = 2`

so from here let's take the natural log

`\text{ln}(e^(0.0525t))=ln(2)`

the expression
`ln(e^(0.0525t))` is essentially saying what do I have to raise the base "e" to the power of, to get "e^0.0525t" and we're essentially given the answer... 0.0525t is what we need to raise it to the power of, so that's what it simplifies to

`0.0525t = ln(2)`

from here divide both sides by 0.0525

`t=(ln(2))/(0.0525)`

now approximate natural log using a calculator

`t\approx (0.69314718056)/(0.0525)`

now approximate this using a calculator once more

`t\approx 13.2028034392 \approx 13.2`