SOLUTION: Please help me :)
1) You invest $18000 at 6% interest per year. Set up an exponential equation and use it to determine how much money you will have after 5 years.
2) The chemical
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Please help me :)
1) You invest $18000 at 6% interest per year. Set up an exponential equation and use it to determine how much money you will have after 5 years.
2) The chemical
Log On
Question 185922: Please help me :)
1) You invest $18000 at 6% interest per year. Set up an exponential equation and use it to determine how much money you will have after 5 years.
2) The chemical element strontium has a half-life of 25 years. If 8g were present initialy, how long will it take for the amount to be reduced to 2g? Set up an exponential equation and solve algebraically.
3) Solve for x: 3^(4x-1)= 9^1/3 Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 1) You invest $18000 at 6% interest per year. Set up an exponential equation
and use it to determine how much money you will have after 5 years.
:
A = 18000*(1+.06)^5
A = 18000*1.06^5
A = 18000*1.3382
A = $24,088.06
;
:
2) The chemical element strontium has a half-life of 25 years.
If 8g were present initially, how long will it take for the amount to be
reduced to 2g? Set up an exponential equation and solve algebraically.
:
t = time in yrs for this to happen = 2
divide both sides by 8 = = .25
Using nat logs *ln(2) = ln(.25)
.693* = -1.386
:
Multiply both sides by 25
-.693t = -1.386 * 25
:
-.693t = -34.657
t =
t = +50.0 yrs
:
:
3) Solve for x: 3^(4x-1)= 9^1/3
Cube both sides and you have;
therefore
12x - 3 = 2
12x = 2 + 3
x =
:
:
you can check all three solutions by substitution and entering the
original equations into a good calc
