Posted by **Phoebe** on Thursday, June 20, 2013 at 5:14pm.

20. Growth of an Account If Russ (see Exercise 19) chooses the plan with continuous compounding, how long will it take for his $60,000 to grow to $80,000?

3. (1/8) to the 2x-3= 16 to the x +1

13. e to the 4x= 4 to the x-2

15. ln (x-4) ln 3= ln(1/5) to the x power

20. Tripling Time For any amount of money invested at 6.8% annual interest compounded continuously, how long will it take to triple?

Please someone help ASAP!!!!!!!!! I don't understand any of these college algebra problems.

- College Algebra -
**Reiny**, Thursday, June 20, 2013 at 5:45pm
19

you want to solve for t

60000 e^.0675t = 80000

e^.0675t = 1.333...

ln both sides and use rules of logs

.0675t lne = ln 1.3333... ( but lne = 1)

t = ln 1.3333../.0675 = 4.26 years, appr 4 years and 3 months

3.

(1/8)^(2x-3) = 16^(x+1)

(2^-3)^(2x-3) = (2^4)^(x+1)

2^(-6x + 9 = 2^(4x + 4)

-6x + 9 = 4x + 4

-10x = -5

x = 1/2

13.

e^(4x) = 4^(x-2)

ln both sides

4x lne = (x-2)ln4

4x = ln4 x - 2ln4

4x - ln4 x = -2ln4

x(4 - ln4) = -2ln4

x = -2ln4/(4-ln4)

= appr -1.0608

15. ln(x-4) ln3 = ln(1/5)^x

no simple algebraic way to solve this, unless I misread the question. I took it the way you typed it

ran it through Wolfram and Wolfram ran out of time trying to solve it

http://www.wolframalpha.com/input/?i=solve++ln%28x-4%29*ln3+%3D+ln%281%2F5%29%5Ex

20. solve

1 e^.068t = 3

now do it the same way as #19

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