exponential growth of investment

Question

Robin has some money invested in an account that pays 6 percent interest per year.

At what rate is the investment increasing each decade?

What is the monthly rate of growth?

Answer 1

(1.06)^10 = about 1.79 which is about a 79% increase in 10 years = 79% per decade

interest compounded annually

use a calculator

(1.06)^(1/12) = about 1.0049 which is about 0.5% or 1/2 of 1% increase per month

or 6%/12 = 6/12 = 1/2% = 0.5% per month

Answer 2

Every year the money increases by a factor of (1+.06)

In ten years the factor will be (1.06)¹⁰

The "monthly rate of growth" is ambiguous to me

It could mean 1/12 the yearly rate of .06 (0.5% or .005)

or it could mean the monthly interest compounded monthly that would yield the same yearly interest(1.06)

(1+i_month)¹² = 1.06 I_month = .00487

Please consider a tutor. Take care.

Answer 3

A(t) = a₀(1.06)^t

A(10) = 1.79a₀

1.79a₀ = a₀e^10k

Can you solve for k and answer?

A(1/2) = a₀(1.06)^1/12

Can you follow A(10) as an example?