i need help with rates

Question

If you had $160,000 in account A growing at a rate of 3% each year, and $1,000 in account B growing at 6% each year, how long would it take for account B to surpass account A?

Answer 1

Let's first model these two as equations:

 

X is the number of years that we let each account grow.

 

A = 160000*(1.03^X)

B = 1000*(1.06^X)

 

Now, we want to know when B is greater than A, so we can solve X for B > A

 

B > A

1000(1.06^X) > 160000(1.03^X)

(1.06^X) > 160(1.03^X) - Just simplifying!

 

Now you will need to freshen up on your log rules, you can just Google "Basic Log Rules" and a bunch should pop up. 

 

Here is a good one: http://www.rapidtables.com/math/algebra/logarithm/Logarithm_Rules.htm

 

LOG(1.06^X) > LOG(160*1.03^X)

X*LOG(1.06) > LOG(160) + XLOG(1.03) 

X*(LOG(1.06) - LOG(1.03)) > LOG(160) 

X > LOG(160) / LOG(1.06/1.03)) - 

X > 177

 

Answer 2

This problem is utilizing the interest rate formula in which is:

 

A = P*(1+r/n)^nt  n=1 due to the compounding occurring once a year  

 

The first step is to set up the formula for each account in which are:

 

A=160,000*(1+0.03/1)^1*t

 

and

 

B=1000*(1+0.06/1)^1*t

 

The key to finding out when account B will surpass account A is to equate the two equations and solve for t.

 

160,000*(1+0.03)^t = 1000*(1+0.06)^t

 

This turns out to be you'll have to use logarithms to solve for t. I'll leave this up to you to come up with the final answer.